3 Marks Question

Question 13 Marks

Differentiate function $\frac{\sin x-x \cos x}{x \sin x+\cos x}$ $\text {w.r.t.x.}$

Answer

Let $y=\frac{\sin x-x \cos x}{x \sin x+\cos x}$
On differenting both sides $\text {w.r.t.x,}$
$(x \sin x+\cos x) \frac{d}{d x}(\sin x-x \cos x)-$
$\therefore \frac{d y}{d x}=\frac{(\sin x-x \cos x) \frac{d}{d x}(x \sin x+\cos x)}{(x \sin x+\cos x)^2}$
$=\frac{\begin{array}{c}(x \sin x+\cos x)(\cos x-\cos x+x \sin x)- \\ (\sin x-x \cos x)(x \cos x+\sin x-\sin x)\end{array}}{(x \sin x+\cos x)^2}$
$\frac{d y}{d x}=\frac{\begin{array}{c}(x \sin x+\cos x)(x \sin x)- \\ (\sin x-x \cos x)(x \cos x)\end{array}}{(x \sin x+\cos x)^2}$
$\frac{d y}{d x}=\frac{x\left(x \sin ^2 x+\sin x \cos x-\sin x \cos x+x \cos ^2 x\right)}{(x \sin x+\cos x)^2}$
$\therefore \frac{d y}{d x}=\frac{x^2\left(\sin ^2 x+\cos ^2 x\right)}{(x \sin x+\cos x)^2}=\frac{x^2}{(x \sin x+\cos x)^2}$

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Question 23 Marks

Calculate differentiation of function $f(x)$ $\frac{e^x+e^{-x}}{e^x-e^{-x}}$ $\text{w.r.t.x.}$

Answer

Let $y=\frac{e^x+e^{-x}}{e^x-e^{-x}}$
On differentiating both sides $\text {w.r.t.x}:$
$\frac{d y}{d x}=\frac{\left(e^x-e^{-x}\right) \frac{d}{d x}\left(e^x+e^{-x}\right)-\left(e^x+e^{-x}\right) \frac{d}{d x}\left(e^x-e^{-x}\right)}{\left(e^x-e^{-x}\right)^2}$
$=\frac{\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)-\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)}{\left(e^x-e^{-x}\right)^2}$
$=\frac{\left(e^x-e^{-x}\right)^2-\left(e^x+e^{-x}\right)^2}{\left(e^x-e^{-x}\right)^2}$
$=\frac{\left(e^x-e^{-x}+e^x+e^{-x}\right)\left(e^x-e^{-x}-e^x-e^{-x}\right)}{\left(e^x-e^{-x}\right)^2}$
$=\frac{2 e^x \times\left(-2 e^{-x}\right)}{\left(e^x-e^{-x}\right)^2}=\frac{-4 e^0}{\left(e^x-e^{-x}\right)^2}=\frac{-4 \times 1}{\left(e^x-e^{-x}\right)^2}$
$=\frac{-4}{\left(e^x-e^{-x}\right)^2}$

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