Question 13 Marks
Differentiate the function $(\sin x – \cos x^{(\sin x – \cos x)},\frac{\pi}{4}$
AnswerWe have,$ y = (\sin x - \cos x)^{(\sin x - \cos x)}, \frac { \pi } { 4 } < x < \frac { 3 \pi } { 4 }$,Therefore, on taking logarithm both sides, we get,
$\log y = \log (\sin x - \cos x)^{(\sin x - \cos x)},$
$ \Rightarrow \log y= (\sin x - \cos x). \log(\sin x - \cos x)$
Therefore,on differentiating both sides w.r.t x, we get,
$ \frac { 1 } { y } \cdot \frac { d y } { d x } = ( \sin x - \cos x ) \times \frac { d } { d x } \log ( \sin x - \cos x ) + \log(\sin x - \cos x) \times\frac { d } { d x } ( \sin x - \cos x )$[By using product rule of derivative]
$ \Rightarrow \quad \frac { 1 } { y } \frac { d y } { d x } = (\sin x - \cos x) \frac { 1 } { ( \sin x - \cos x ) }$$ \frac { d } { d x } ( \sin x - \cos x )+ \log(\sin x - \cos x).(\cos x + \sin x)$
$ \Rightarrow \quad \frac { 1 } { y } \frac { d y } { d x } = ( \sin x - \cos x ) \frac { 1 } { ( \sin x - \cos x ) }(\cos x + \sin x) + \log(\sin x - \cos x).(\cos x + \sin x)$
$ \Rightarrow \quad \frac { 1 } { y } \frac { d y } { d x } = ( \cos x + \sin x )+(\cos x + \sin x) + \log(\sin x - \cos x)$
$ \Rightarrow \quad \frac { d y } { d x }= y(\cos x + \sin x)[ 1+\log(\sin x - \cos x)]$
$ \therefore \quad \frac { d y } { d x } = (\sin x - \cos x)^{(\sin x - \cos x)}(\cos x + \sin x)[ 1+\log(\sin x - \cos x)]$
View full question & answer→Question 23 Marks
Differentiate the function $(\log x)^{\log x}, x > 1,$ w.r.t to x.
AnswerLet $y = (\log x)^{\log x}, x > 1$
Taking logarithm on both sides
$\Rightarrow \log y = \log (\log x)^{\log x} = \log x \times \log (\log x)$
Differentiating both sides with respect to $x,$ we get
$\frac{1}{\mathrm{y}} \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\log \mathrm{x} \times \log (\log \mathrm{x})]$
$\Rightarrow$ $\frac{1}{y} \frac{d y}{d x}=\log (\log x) \times \frac{d}{d x}(\log x)+\log x \times \frac{d}{d x}[\log (\log x)]$
$\Rightarrow \frac{d y}{d x}=y\left[\log (\log x) \times \frac{1}{x}+\log x \times \frac{1}{\log x} \times \frac{d}{d x}(\log x)\right]$
$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x} \log (\log x)+\frac{1}{x}\right]$
$\therefore \frac{d y}{d x}=(\log x)^{\log x}\left[\frac{1}{x}+\frac{\log (\log x)}{x}\right]$
View full question & answer→Question 33 Marks
Differentiate the function $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}$ w.r.t. x.
Answer$y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} + \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} - \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}} \right]$ $\left[ \because\sqrt {1 \pm \sin x} = \sqrt {{{\left( {\cos \frac{x}{2} \pm \sin \frac{x}{2}} \right)}^2}} \right] $ $= {\cot ^{ - 1}}\frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2}}}$
$ = {\cot ^{ - 1}}\left( {\frac{{2\cos \frac{x}{2}}}{{2\sin \frac{x}{2}}}} \right)$
$ = {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right)$
$ = \frac{x}{2}$
$y = \frac{x}{2}$
$\frac{{dy}}{{dx}} = \frac{1}{2}$
View full question & answer→Question 43 Marks
Differentiate the function $\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2<x<2$ w.r.t to x.
AnswerLet y = $\frac{\cos ^{-1} \frac x2}{\sqrt{2 x+7}},-2<x<2$
Differentiating both sides with respect to x, we get
Using Quotient rule
$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}\right)-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}}$
$\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}$ = $\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \times \frac{d}{d x}\left(\frac{x}{2}\right)\right]-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{1}{2 \sqrt{2 x+7}} \times \frac{d}{d x}(2 x+7)}{2 x+7}$
$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \times-\frac{1}{\sqrt{4-x^{2}}}-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{2}{2 \sqrt{2 x+7}}}{2 x+7}$
$\frac{d y}{d x}=-\frac{\sqrt{2 x+7}}{\sqrt{4-x^{2}} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)}$
$\therefore$ $\frac{d y}{d x}=-\left[\frac{1}{\sqrt{4-x^{2}} \times \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right]$
View full question & answer→Question 53 Marks
Differentiate the function $sin ^{-1}({x\sqrt x})\ ,{0 \leq x \leq 1}$ w.r.t. to x.
AnswerLet $y = {\sin ^{ - 1}}\left( {x\sqrt x } \right) = {\sin ^{ - 1}}\left( {{x^{\frac{3}{2}}}} \right)$ $\therefore \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - \left( {{x^{\frac{3}{2}}}} \right)^2} }}\frac{d}{{dx}}{x^{\frac{3}{2}}}$
$= \frac{1}{{\sqrt {1 - {x^3}} }}.\frac{3}{2}{x^{\frac{1}{2}}}$
$= \frac{{3\sqrt x }}{{2\sqrt {1 - {x^3}} }}$
$= \frac{3}{2}\sqrt {\frac{x}{{1 - {x^3}}}}$
View full question & answer→Question 63 Marks
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
AnswerConsider the function f (x) = |x| + |x - 1|
f is continuous everywhere but it is not differentiable at x = 0 and x = 1.
View full question & answer→Question 73 Marks
Using the fact that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and the differentiation, obtain the sum formula for cosines.
AnswerGiven: $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ Consider, A and B as function of t and differentiating both sides w.r.t. x,
$\cos \left( {A + B} \right)\left( {\frac{{dA}}{{dt}} + \frac{{dB}}{{dt}}} \right) = \sin A\left( { - \sin B} \right)\frac{{dB}}{{dt}} + \cos B\left( {\cos A\frac{{dA}}{{dt}}} \right)$$ + \cos A\cos B\frac{{dB}}{{dt}} + \sin B\left( { - \sin A} \right)\frac{{dA}}{{dt}}$
$\Rightarrow \cos \left( {A + B} \right)\left( {\frac{{dA}}{{dB}} + \frac{{dB}}{{dt}}} \right) = \left( {\cos A\cos B - \sin A\sin B} \right)\left( {\frac{{dA}}{{dt}} + \frac{{dB}}{{dt}}} \right)$
$\Rightarrow \cos \left( {A + B} \right) = \left( {\cos A\cos B - \sin A\sin B} \right)$
View full question & answer→Question 83 Marks
If $f(x) = | x |^3$ , show that f ″(x) exists for all real x and find it.
AnswerWe know that,
$|x|=\left\{\begin{array}{r} {x, \text { if } x \geq 0} \\ {-x, \text { if } x<0} \end{array}\right.$
When, x $\ge$ 0,
Then $f(x) = |x|^3 = x^3$
$\implies f’(x) = 3x^2$
And $f^{\prime\prime}(x)=6x$
When x < 0,
Then, $f(x) = |x|^3 = ( – x)^3 = – x^3$
$\implies$$f^\prime(x)=-3x^2$,
And, $f^{\prime\prime}(x)=-6x$
$\therefore$ $f^{\prime\prime}(x)$ = $\left\{\begin{array}{r} {6 x, x \geq 0} \\ {-6 x, x<0} \end{array}\right.$
View full question & answer→Question 93 Marks
If x = a (cos t + t sin t) and y = a (sin t – t cos t), find $\frac{{{d^2}y}}{{d{x^2}}}$
Answerx = a(cos t + t.sin t) $\frac{{dy}}{{dt}} = a\left[ { - \sin t + t\cos t + \sin t.1} \right]$
$\frac{{dy}}{{dt}} = a\left( {t.\cos t} \right)$ ...(1)
y = a(sin t - tcos t)
$\frac{{dy}}{{dt}} = a\left[ {\cos t - \left( { - t\sin t + \cos t.1} \right)} \right]$
= a( cos t +t sin t - cos t)
= a(t sin t)
$\frac{{dy}}{{dx}} = \frac{{a\,t.\sin t}}{{a\,t.\cos t}} = \tan t$
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\tan t} \right)$
$= \frac{d}{{dt}}\left( {\tan t.} \right)\frac{{dt}}{{dx}}$
$= {\sec ^2}t.\frac{1}{{a\,t.\cos t}}$
$= \frac{1}{{a\,t.{{\cos }^3}t}}$
View full question & answer→Question 103 Marks
If $\cos y = x\cos \left( {a + y} \right)$ with $\cos a \ne \pm 1$ prove that $\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}$
AnswerGiven: $\cos y = x\cos \left( {a + y} \right)$ $\Rightarrow x = \frac{{\cos y}}{{\cos \left( {a + y} \right)}}$
$\therefore \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left( {\frac{{\cos y}}{{\cos \left( {a + y} \right)}}} \right)$
$\Rightarrow \frac{{dx}}{{dy}} = \frac{{\cos \left( {a + y} \right)\frac{d}{{dy}}\cos y - \cos y\frac{d}{{dy}}\cos \left( {a + y} \right)}}{{{{\cos }^2}\left( {a + y} \right)}}$
$ \Rightarrow \frac{{dx}}{{dy}} = \frac{{\cos \left( {a + y} \right)\left( { - \sin y} \right) - \cos y\left\{ { - \sin \left( {a + y} \right)} \right\}}}{{{{\cos }^2}\left( {a + y} \right)}}$
$= \frac{{ - \cos \left( {a + y} \right)\sin y + sin\left( {a + y} \right)\cos y}}{{{{\cos }^2}\left( {a + y} \right)}}$
$\Rightarrow \frac{{dx}}{{dy}} = \frac{{\sin \left( {a + y - y} \right)}}{{{{\cos }^2}\left( {a + y} \right)}}$
$= \frac{{\sin a}}{{{{\cos }^2}\left( {a + y} \right)}}$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}$ [ reciprocal]
View full question & answer→Question 113 Marks
If $x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ for -1<x<1 prove that $\frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {1 + x} \right)}^2}}}$
Answer$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ $x\sqrt {1 + y} = - y\sqrt {1 + x} $
Squaring both sides,
${x^2}\left( {1 + y} \right) = {y^2}\left( {1 + x} \right)$
${x^2} + {x^2}y = {y^2} + x{y^2}$
${x^2} - {y^2} + {x^2}y - x{y^2} = 0$
$\left( {x - y} \right)\left( {x + y} \right) + xy\left( {x - y} \right) = 0$
$\left( {x - y} \right)\left[ {x + y + xy} \right] = 0$
$x + y + xy = 0$
$y\left( {1 + x} \right) = - x$
$y = \frac{{ - x}}{{1 + x}}$
$\frac{{dy}}{{dx}} = - \left[ {\frac{{\left( {1 + x} \right)\left( 1 \right) - \left( x \right)\left( 1 \right)}}{{{{\left( {1 + x} \right)}^2}}}} \right]$$= - \left[ {\frac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}}} \right]$
$\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}$
View full question & answer→Question 123 Marks
Find $\frac{d y}{d x}$, if $y = \sin^{–1} x + \sin^{–1} \sqrt{1-x^{2}}, 0 < x < 1$
Answer$y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}} $
Differentiate both side w.r.t. x
$\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - \left( {\sqrt {1 - {x^2}} } \right)^2} }}\frac{d}{{dx}}\left( {\sqrt {1 - {x^2}} } \right)$
$=\frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - \left( {\sqrt {1 - {x^2}} } \right)^2} }}.\frac{{1\left( { - 2x} \right)}}{{2\sqrt {1 - {x^2}} }}$
$= \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - \left( {\sqrt {1 - {x^2}} } \right)^2} }}.\left( {\frac{{ - x}}{{\sqrt {1 - {x^2}} }}} \right)$
$= \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {{x^2}} }}.\left( {\frac{{ - x}}{{\sqrt {1 - {x^2}} }}} \right)$
$= \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{x}\left( {\frac{{ - x}}{{\sqrt {1 - {x^2}} }}} \right)$
$= \frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {x^2}} }} = 0$
View full question & answer→Question 133 Marks
Differentiate w.r.t. $x$ the function in $x^x + x^a + a^x + a^a ,$ for some fixed $a > 0$ and $x > 0.$
AnswerLet $y = x^x + x^a + a^x + a^a,$ for some fixed a > 0 and $x > 0$
And let $x^x = u, x^a = v a^x = w$ and $a^a = s$
Then $y = u + v + w + s$
$\therefore$ $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x} ......(i)$
Now,
$u = x^x$
Taking logarithm both sides, we get
$\log u = \log x^x$
$\Rightarrow \log u = x \log x$
Differentiating both sides w.r.t. $x$
$\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\log \mathrm{x} \times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\mathrm{x} \times \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d u}{d x}=u\left[\log x+x \times \frac{1}{x}\right]$
$\Rightarrow \frac{d u}{d x}=x^{x}[\log x+1]=x^{x}(1+\log x) ......(ii)$
$v = x^a$
Differentiating both sides with respect to $x$
$\frac{d v}{d x}=\frac{d}{d x}\left(x^{a}\right)$
$\Rightarrow \frac{d v}{d x}=a x^{a-1} .....(iii)$
$w = a^x$
Taking logarithm both sides
$\log w = \log a^x$
$\log w = x \log a$
Differentiating both sides with respect to $x$
$\frac{1}{w} \frac{d w}{d x}=\log a \times \frac{d}{d x}(x)$
$\Rightarrow \frac{d w}{d x}=w \log a$
$\Rightarrow \frac{d w}{d x}=a^{x} \log a ......(iv)$
$s = a^a$
Differentiating both sides with respect to $x$
$\frac{\mathrm{ds}}{\mathrm{dx}}=0 .....(v)$
Putting $(ii), (iii), (iv)$ and $(v)$ in $(i)$
$\frac{d y}{d x}=x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a+0$
$\therefore \frac{d y}{d x}=x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$
View full question & answer→Question 143 Marks
Find the second-order derivative of the function log(log x)
AnswerLet y = log(log x) $\therefore \frac{{dy}}{{dx}} = \frac{1}{{\log x}}\frac{d}{{dx}}\log x\left[ {\therefore \frac{d}{{dx}}\log f\left( x \right) = \frac{1}{{f\left( x \right)}}\frac{d}{{dx}}f\left( x \right)} \right]$
$ = \frac{1}{{\log x}}.\frac{1}{x} = \frac{1}{{x\log x}}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\left( {x\log x} \right)\frac{d}{{dx}}\left( 1 \right) - 1\frac{d}{{dx}}\left( {x\log x} \right)}}{{{{\left( {x\log x} \right)}^2}}}$
$= \frac{{\left( {x\log x} \right)\left( 0 \right) - \left[ {x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}x} \right]}}{{{{\left( {x\log x} \right)}^2}}}$
$= \frac{{ - \left[ {x.\frac{1}{x} + \log x \times 1} \right]}}{{{{\left( {x\log x} \right)}^2}}}$
$= \frac{{ - \left[ {1 + \log x} \right]}}{{{{\left( {x\log x} \right)}^2}}}$
View full question & answer→Question 153 Marks
Find the second-order derivatives of the function $\tan^{-1}x$
AnswerLet $y = \tan^{-1}x$
$\therefore \frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{1}{{1 + {x^2}}}} \right)$
$= \frac{{\left( {1 + {x^2}} \right)\frac{d}{{dx}}\left( 1 \right) - 1\frac{d}{{dx}}\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}$
$= \frac{{\left( {1 + {x^2}} \right) \times 0 - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}$
View full question & answer→Question 163 Marks
Find the second-order derivatives of the function $e^{6x}\cos 3x$
AnswerLet $y = e^{6x}\cos 3x \therefore \frac{{dy}}{{dx}} = {e^{6x}}.\frac{d}{{dx}}\cos 3x + \cos 3x\frac{d}{{dx}}{e^{6x}}$
$= {e^{6x}}\left( { - \sin 3x} \right)\frac{d}{{dx}}\left( {3x} \right) + \cos 3x.{e^{6x}}\frac{d}{{dx}}\left( {6x} \right)$
$= -{e^{6x}}\sin 3x \times 3 + \cos 3x.{e^{6x}} \times 6$
$= e^{6x}(-3 \sin 3x + 6 \cos 3x)$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = {e^{6x}}\frac{d}{{dx}}\left( { - 3\sin 3x + 6\cos 3x} \right) $ $+ \left( { - 3\sin 3x + 6\cos 3x} \right)\frac{d}{{dx}}{e^{6x}}$
$= {e^{6x}}\left( { - 3\cos 3x \times 3 - 6\sin 3x \times 3} \right) $ $+ \left( { - 3\sin 3x + 6\cos 3x} \right){e^{6x}} \times 6$
$= e^{6x}(-9 \cos 3x - 18 \sin 3x - 18 \sin 3x + 36 \cos 3x)$
$= e^{6x}(27 \cos 3x - 36 \sin 3x)$
$= 9e^{6x}(3 \cos 3x - 4 \sin 3x)$
View full question & answer→Question 173 Marks
Find the second-order derivative of the function $e^x\sin 5x$
AnswerLet $y = e^x \sin 5x \therefore \frac{{dy}}{{dx}} = {e^x}\frac{d}{{dx}}\sin 5x + \sin 5x\frac{d}{{dx}}{e^x}$
$= {e^x}\cos 5x\frac{d}{{dx}}5x + \sin 5x.{e^x} = {e^x}\cos 5x \times 5 + {e^x}\sin 5x$
$= e^x(5 \cos 5x + \sin 5x)$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = {e^x}\frac{d}{{dx}}\left( {5\cos 5x + \sin 5x} \right) + \left( {5\cos 5x + \sin 5x} \right)\frac{d}{{dx}}{e^x}$
$= {e^x}\left[ {5\left( { - \sin 5x} \right) \times 5 + \left( {\cos 5x} \right) \times 5} \right] + (5 \cos 5x + \sin 5x)e^x$
$= e^x(-25 \sin 5x + 5 \cos 5x + 5 \cos 5x + \sin 5x)$
$= e^x(10 \cos 5x - 24 \sin 5x)$
$= 2e^x(5 \cos 5x - 12 \sin 5x)$
View full question & answer→Question 183 Marks
Find the second-order derivative of the function $x^3$ log $x$
AnswerLet $y = x^3$ log x$
$\therefore \frac{{dy}}{{dx}} = {x^3}\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}{x^3}$
$= {x^3}.\frac{1}{x} + \log x\left( {3{x^2}} \right)$
$= x^2 + 3x^2log x$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {{x^2} + 3{x^2}\log x} \right)$
$= \frac{d}{{dx}}{x^2} + 3\frac{d}{{dx}}\left( {{x^2}\log x} \right)$
$= 2x + 3\left[ {{x^2}\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}{x^2}} \right]$
$= 2x + 3\left( {{x^2}.\frac{1}{x} + \left( {\log x} \right)2x} \right)$
$= 2x + 3(x + 2x \log x)$
$= 2x + 3x + 6x \log x$
$= 5x + 6x \log x$
$= x(5 + 6 \log x)$
View full question & answer→Question 193 Marks
If $y = \cos^{-1}x$. Find $\frac{{{d^2}y}}{{d{x^2}}}$ in terms of y alone.
AnswerGiven: $y = \cos^{-1}x$
$\Rightarrow x = \cos y$ .....(i)
$\therefore \frac{{dy}}{{dx}} = \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}$
$= \frac{{ - 1}}{{\sqrt {1 - {{\cos }^2}y} }}$ [From eq. (i)]
$ = \frac{{ - 1}}{{\sqrt {{{\sin }^2}y} }} = \frac{{ - 1}}{{\sin y}} = - {\text{cosec}}\,y$ .....(ii)
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = - \frac{d}{{dx}}\left( {\text cosec\,y} \right)$
$= - \left[ { - \ cosec\ y\,\cot y\frac{{dy}}{{dx}}} \right]$
$= cosecy \cot y(-cosec y) [ using (ii)]$
$= -cosec^2y\ coty$
View full question & answer→Question 203 Marks
If y = 5 cos x - 3 sin x, prove that $\frac{{{d^2}y}}{{d{x^2}}} + y = 0$
AnswerLet y = 5 cos x - 3 sin x ....(i) $\therefore \frac{{dy}}{{dx}} = - 5\sin x - 3\cos x$
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = - 5\cos x + 3\sin x$
= -(5 cos x - 3 sin x) = -y [From eq. (i)]
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = - y$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} + y = 0$
View full question & answer→Question 213 Marks
Find the second-order derivative of the function sin(log x)
AnswerLet y = sin (log x) $\therefore \frac{{dy}}{{dx}} = \cos \left( {\log x} \right)\frac{d}{{dx}}\left( {\log x} \right)$
$= \cos \left( {\log x} \right).\frac{1}{x} = \frac{{\cos \left( {\log x} \right)}}{x}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{x\frac{d}{{dx}}\cos \left( {\log x} \right) - \cos \left( {\log x} \right)\frac{d}{{dx}}x}}{{{x^2}}}$
$= \frac{{x (-sinlogx)\frac{d}{{dx}}\left( {\log x} \right) - \cos \left( {\log x} \right) \times 1}}{{{x^2}}}$
$= \frac{{ - x\sin \left( {\log x} \right)\frac{1}{x} - \cos \left( {\log x} \right)}}{{{x^2}}}$
$= \frac{{ - \left[ {\sin \left( {\log x} \right) + \cos \left( {\log x } \right)} \right]}}{{{x^2}}}$
View full question & answer→Question 223 Marks
If x and y are connected parametrically by the equation $x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),$ y = a sin t, without eliminating the parameter, find $\frac{{dy}}{{dx}}$.
AnswerGiven: $x = a\left( {\cos t + \log \tan \frac{t}{2}} \right)$ and y = a sin t
$\therefore \frac{{dx}}{{dt}} = a\left[ { - \sin t + \frac{1}{{\tan \frac{t}{2}}}.\frac{d}{{dt}}\left( {\tan \frac{t}{2}} \right)} \right]$
$= a\left[ { - \sin t + \frac{1}{{\tan \frac{t}{2}}}.{{\sec }^2}\frac{t}{2}.\frac{1}{2}} \right]$
$= a\left[ { - \sin t + \frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}}.\frac{1}{{{{\cos }^2}\frac{t}{2}}}.\frac{1}{2}} \right]$
$= a\left[ { - \sin t + \frac{1}{{2\sin \frac{t}{2}\cos \frac{t}{2}}}} \right]$
$= a\left[ { - \sin t + \frac{1}{{\sin t}}} \right]$
$ = a\left( {\frac{{1 - {{\sin }^2}t}}{{\sin t}}} \right) = \frac{{a{{\cos }^2}t}}{{\sin t}}$
And $\frac{{dy}}{{dt}} = a\cos t$
$\therefore \frac{{dy}}{{dt}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{a\cos t}}{{\left( {\frac{{a{{\cos }^2}t}}{{\sin t}}} \right)}} = \frac{{\sin t}}{{\cos t}} = \tan t$
View full question & answer→Question 233 Marks
If x and y are connected parametrically by the equation $x = a\left( {\cos \theta + \theta \sin \theta } \right),$ $y = a\left( {\sin \theta - \theta \cos \theta } \right)$, without eliminating the parameter, find $\frac{{dy}}{{dx}}$.
AnswerGiven: $x = a\left( {\cos \theta + \theta \sin \theta } \right)$ and $y = a\left( {\sin \theta - \theta \cos \theta } \right)$
$\therefore \frac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \theta \cos \theta + \sin \theta .1} \right)$
$ = a\theta \cos \theta$
And $\frac{{dy}}{{d\theta }} = a\left[ {\cos \theta - \left\{ {\theta \left( { - \sin \theta } \right) + \cos \theta .1} \right\}} \right]$
$= a\left[ {\cos \theta + \theta \sin \theta - \cos \theta } \right] = a\theta \sin \theta$
$\therefore \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{a\theta \sin \theta }}{{a\theta \cos \theta }} = \tan \theta$
View full question & answer→Question 243 Marks
Differentiate the function $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$ w.r.t. x.
AnswerGiven: $(x + 3)^2.(x + 4)^3.(x + 5)^4$
Let y = $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$
Taking log on both sides, we get
log y = $\log \left((x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}\right)$
$\Rightarrow \log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4}$
$\Rightarrow \log y=2 \cdot \log (x+3)+3 \cdot \log (x+4)+4 \cdot \log (x+5)$
Now, differentiating both sides with respect to x
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}}(2 \cdot \log (\mathrm{x}+3))+\frac{\mathrm{d}}{\mathrm{dx}}(3 \cdot \log (\mathrm{x}+4))+\frac{\mathrm{d}}{\mathrm{dx}}(4 \cdot \log (\mathrm{x}+5))$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)$
$\Rightarrow \frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$
$\Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^4$ $\left[\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]$
$\begin{aligned} \Rightarrow \frac{\mathrm{d} y}{\mathrm{dx}}=(\mathrm{x}+&3)^{1}(\mathrm{x}+4)^{2}(\mathrm{x}+5)^{3}\left[2\left(\mathrm{x}^{2}+9 \mathrm{x}+20\right)+3\left(\mathrm{x}^{2}+8 \mathrm{x}+15\right)\right. \left.+4\left(\mathrm{x}^{2}+7 \mathrm{x}+12\right)\right] \end{aligned}$
$=$ $(x+3)(x+4)^{2}(x+5)^{3}\left(9 x^{2}+70 x+133\right)$
View full question & answer→Question 253 Marks
Differentiate the function $x^{x}-2^{\sin x}$ w.r.t. x.
AnswerGiven: $x^x - 2 \sin x$
Let $y = x^x - 2 \sin x$
Let $y = u - v$
$\Rightarrow u = x^x$ and $v = 2 \sin x$
For, $u = x^x$
Taking \log on both sides, we get
$\log u = \log x^x$
$\Rightarrow \log u = x. \log(x)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\mathrm{x})]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\mathrm{x}}+\log \mathrm{x} \cdot(1)\right]$
$\Rightarrow \frac{d u}{d x}=x^{x}(1+\log x)$
For, $v = 2^{\sin x}$
Taking \log on both sides, we get
$\log v = \log 2^{\sin x}$
$\Rightarrow \log v = \sin x. \log (2)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{v})=\frac{\mathrm{d}}{\mathrm{dx}}[\sin \mathrm{x} \cdot \log (2)]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\log 2 \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})$
$\Rightarrow \frac{d v}{d x}=v[\log 2 .(\cos x)]$
$\Rightarrow \frac{d v}{d x}=2^{\sin x} \cdot \cos x \cdot \log 2$
Because,$y = u - v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$
$\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x }\cos x \log 2$
View full question & answer→Question 263 Marks
Differentiate the function $\sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} $ w.r.t. x.
AnswerLet $y = \sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} = {\left( {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} \right)^{\frac{1}{2}}}$ ……….(i) Taking log on both sides, we have
$ \log y = \frac{1}{2}\left[ {\log \left( {x - 1} \right) + \log \left( {x - 2} \right) - \log \left( {x - 3} \right) - \log \left( {x - 4} \right) - \log \left( {x - 5} \right)} \right]$
Differentiating both sides w.r.t x, we get
$\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{1}{{x - 1}}\frac{d}{{dx}}\left( {x - 1} \right) + \frac{1}{{x - 2}}\frac{d}{{dx}}\left( {x - 2} \right) - \frac{1}{{x - 3}}\frac{d}{{dx}}\left( {x - 3} \right) - \frac{1}{{x - 4}}\frac{d}{{dx}}\left( {x - 4} \right) - \frac{1}{{x - 5}}\frac{d}{{dx}}\left( {x - 5} \right)} \right]$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{2}y\left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} -\frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{2}\sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} \left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$
View full question & answer→Question 273 Marks
Differentiate the function $\cos x \cdot \cos 2 x \cdot \cos 3 x$ w.r.t. x.
AnswerGiven function is: $cosx.cos2.cos3x$
Let y = $cosx.cos2.cos3x$
Taking log on both sides, we get
log y = log(cos x. cos 2x. cos 3x)
Now, differentiate both sides with respect to x
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}} \log (\cos \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}} \log (\cos 2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(\log \cos 3 \mathrm{x})$
$\implies$ $\frac{1}{\mathrm{y}} \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{1}{\cos \mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\cos \mathrm{x})+\frac{1}{\cos 2 \mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\cos 2 \mathrm{x})+\frac{1}{\cos 3 \mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\cos 3 \mathrm{x})$
$\implies$ $\frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)\right]$
$\implies$ $\frac{\mathrm{dy}}{\mathrm{dx}}=-\cos \mathrm{x} \cdot \cos 2 \mathrm{x} \cdot \cos 3 \mathrm{x}[\tan \mathrm{x}+\tan 2 \mathrm{x}(2)+\tan 3 \mathrm{x}(3)]$
$\implies$ $\frac{\mathrm{dy}}{\mathrm{dx}}=-\cos \mathrm{x} \cdot \cos 2 \mathrm{x} \cdot \cos 3 \mathrm{x}[\tan \mathrm{x}+2 \tan 2 \mathrm{x}+3 \tan 3 \mathrm{x}]$
View full question & answer→Question 283 Marks
Prove that the function f given by f(x) = |x - 1|, x $\in$ R is not differentiable at x = 1.
AnswerGiven: f(x) = |x - 1|, x $\in$ R
We know that a function f is differentiable at a point x = c in its domain if both its limits as:
$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f(c + h) - f(c)}}{h}$ are finite and equal.
Now, to check the differentiability of the given function at x = 1,
Let us consider the left-hand limit/ left hand derivative of function f at x = 1, we have
$Lf^\prime(1)$ = $\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f(1 + h) - f(1)}}{h}$
$ = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{|1 + h - 1| - |1 - 1|}}{h}$
$ = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{|h| - 0}}{h}$ $(~\because$ $h < 0, |h| = -h$$)$
$= -1$
Now, let we consider the right hand limit/right hand derivative of function f at x = 1
$Rf^\prime(1)= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{[{\text{f}}(1 + {\text{h}}) - {\text{f}}(1)]}}{{\text{h}}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{[|1 + h - 1| - |1 - 1|]}}{h}$
$= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{[|h| - 0]}}{h}$
$ = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{[h]}}{h}$ because, { h > 0 $\Rightarrow$ |h| = h}
= 1
Because, left hand limit/left hand derivative is not equal to right hand limit/ right derivative of function f at x = 1, are not equal. So, f is not differentiable at x = 1.
View full question & answer→Question 293 Marks
Differentiate the function with respect to x : $2 \sqrt{\cot \left(x^{2}\right)}$
AnswerGiven function is: $2 \sqrt{\cot \left(x^{2}\right)}$
Let y = $2 \sqrt{\cot \left(x^{2}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(2 \sqrt{\cot \left(x^{2}\right)})$
we know that
$\frac{d}{d x}(k f(x))$ = $k \frac{d}{d x} f(x), \frac{d}{d x}(\sqrt{f(x)})$ = $\frac{1 \ \ \ \ \ \ \ \ \ \ d}{2 \sqrt{f(x)} d x}(f(x))$
Applying both the formula, we get,
$\frac{dy}{dx}=2 \cdot \frac{1}{2 \sqrt{\cot \left(\mathrm{x}^{2}\right)}} \cdot \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}} \cot \left(\mathrm{x}^{2}\right)$
Now, $\frac{d}{d x}(\cot x)=-cosec ^{2} x$
Therefore,
$\frac{dy}{dx}=\frac{1}{\sqrt{\cot \left(x^{2}\right)}} \cdot\left[-\ cosec ^{2}\left(x^{2}\right)\right] \cdot \frac{d}{d x}\left(x^{2}\right)$
= $\sqrt{\frac{\sin x^{2}}{\cos x^{2}}} \times\left(-\frac{1}{\sin ^{2}\left(x^{2}\right)}\right) \times(2 x)$
= $\frac{-2 x}{\sin x^{2} \sqrt{\sin x^{2} \cos x^{2}}}$
= $\frac{-2 x}{\sin x \sqrt[2]{\sin x^{2} \cos x^{2}}} \times \frac{\sqrt{2}}{\sqrt{2}}$
= $\frac{-2 \sqrt{2} x}{\sin x \sqrt[2]{2 \sin x^{2} \cos x^{2}}}$ [Using sin 2x = 2 sin x cos x]
= $\frac{-2 \sqrt{2} x}{\sin x^{2} \sqrt{\sin 2 x^{2}}}$
View full question & answer→Question 303 Marks
Differentiate the function with respect to x : $\cos {x^3}{\sin ^2}\left( {{x^5}} \right)$
AnswerLet $y = \cos {x^3}.{\sin ^2}\left( {{x^5}} \right)$ $\therefore \frac{{dy}}{{dx}} = \cos {x^3}\frac{d}{{dx}}{\sin ^2}\left( {{x^5}} \right) + {\sin ^2}\left( {{x^5}} \right)\frac{d}{{dx}}\cos {x^3}$
$= \cos {x^3}.2\sin \left( {{x^5}} \right)\frac{d}{{dx}}\sin \left( {{x^5}} \right) + {\sin ^2}\left( {{x^5}} \right)\left( { - \sin {x^3}} \right)\frac{d}{{dx}}{x^3}$
$= \cos {x^3}.2\sin \left( {{x^5}} \right)\frac{d}{{dx}}\sin \left( {{x^5}} \right) + {\sin ^2}\left( {{x^5}} \right)\left( { - \sin {x^3}} \right)3{x^2}$
$= \cos {x^3}.2\sin \left( {{x^5}} \right)\cos \left( {{x^5}} \right)\left( {5{x^4}} \right) - {\sin ^2}\left( {{x^5}} \right)\sin {x^3}.3{x^2}$
$= 10{x^4}\cos {x^3}\sin \left( {{x^5}} \right)\cos \left( {{x^5}} \right) - 3{x^2}{\sin ^2}\left( {{x^5}} \right)\sin {x^3}$
View full question & answer→Question 313 Marks
Differentiate the function with respect to x : $\frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}$
AnswerLet $y = \frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}$ Using quotient rule,
$\therefore \frac{{dy}}{{dx}} = \frac{{\cos \left( {cx + d} \right)\frac{d}{{dx}}\sin \left( {ax + b} \right) - \sin \left( {ax + b} \right)\frac{d}{{dx}}\cos \left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= \frac{{\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)\frac{d}{{dx}}\left( {ax + b} \right) - \sin \left( {ax + b} \right)\left\{ { - \sin \left( {cx + d} \right)} \right\}\frac{d}{{dx}}\left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= \frac{{\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)\left( a \right) + \sin \left( {ax + b} \right)\sin \left( {cx + d} \right)\left( c \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= \frac{{a.\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}} + \frac{{c.\sin \left( {ax + b} \right)\sin \left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= a.\cos \left( {ax + b} \right).\sec \left( {cx + d} \right) + c.\sin \left( {ax + b} \right).\tan \left( {cx + d} \right).\sec \left( {cx + d} \right)$
View full question & answer→Question 323 Marks
Prove that the greatest integer function defined by $f\left( x \right) = \left[ x \right],0 < x < 3$ is not differentiable at x = 1 and x = 2.
AnswerGiven: $f\left( x \right) = \left[ x \right],0 < x < 3$
$Rf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}$
= $\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {1 + h} \right] - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 1}}{h}$
= $\mathop {\lim }\limits_{h \to 0} \frac{0}{h}$ = 0
And $Lf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}}$ $= \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {1 - h} \right] - 1}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{0 - 1}}{{ - h}} = \infty $
Since $Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$
Therefore, $f\left( x \right) = \left[ x \right]$ is not differentiable at x =1.
View full question & answer→Question 333 Marks
Differentiate $\sqrt{\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}}$ w.r.t. x.
AnswerLet y = $\sqrt{\frac{(x-3)\left(x^{2}+4\right)}{\left(3 x^{2}+4 x+5\right)}}$
Taking logarithm on both sides, we have
log y = $\frac{1}{2}$[log $(x – 3) + \log (x^2 + 4) – \log (3x^2 + 4x + 5)$]
Now, differentiating both sides w.r.t. x, we get
$\frac{1}{y} \cdot \frac{d y}{d x}$ = $\frac{1}{2}\left[\frac{1}{(x-3)}+\frac{2 x}{x^{2}+4}-\frac{6 x+4}{3 x^{2}+4 x+5}\right]$
or $\frac{d y}{d x}$ = $\frac{y}{2}\left[\frac{1}{(x-3)}+\frac{2 x}{x^{2}+4}-\frac{6 x+4}{3 x^{2}+4 x+5}\right]$
$\Rightarrow$ $~\frac{dy}{dx}$ = $\frac{1}{2} \sqrt{\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}}\left[\frac{1}{(x-3)}+\frac{2 x}{x^{2}+4}-\frac{6 x+4}{3 x^{2}+4 x+5}\right]$
View full question & answer→