3 Marks Question

Question 13 Marks

Find the derivative of function cosec x cot x.

Answer

Here f (x) = cosec x cot x
$\therefore $f'(x) = $\frac{d}{{dx}}$ [cosec x cot x]
= cosec x $\frac{d}{{dx}}$ (cot x) + cot x $\frac{d}{{dx}}$ (cosec x)
= cosec x . – cosec² x + cot x . – cosec x cot x
= - cosec³ x – cosec x cot² x.

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Question 23 Marks

Find the derivative of $\frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$

Answer

Here $f(x) = \frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}} \right]$$= \frac{d}{{dx}}\left( {\frac{2}{{x + 1}}} \right) - \frac{d}{{dx}}\left( {\frac{{{x^2}}}{{3x - 1}}} \right)$
$= \frac{{(x + 1)\frac{d}{{dx}}(2) - 2\frac{d}{{dx}}(x + 1)}}{{{{(x + 1)}^2}}}$$- \frac{{(3x - 1)\frac{d}{{dx}}({x^2}) - {x^2}\frac{d}{{dx}}(3x - 1)}}{{{{(3x - 1)}^2}}}$
$ = \frac{{(x + 1) \times 0 - 2 \times 1}}{{{{(x + 1)}^2}}} - \frac{{(3x - 1)(2x) - {x^2 } \times 3}}{{{{(3x - 1)}^2}}}$
$= \frac{{ - 2}}{{{{(x + 1)}^2}}} - \frac{{6{x^2} - 2x - 3{x^2}}}{{{{(3x - 1)}^2}}}$$ = \frac{{ - 2}}{{{{(x + 1)}^2}}} - \frac{{3{x^2} - 2x}}{{{{(3x - 1)}^2}}}$

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Question 33 Marks

Find the derivative of function 2tan x - 7 sec x

Answer

Here f(x) = 2 tan x - 7 sec x
$\therefore {\text{f}}(x) = \frac{d}{{dx}}[2\tan x - 7\sec x]$
$= 2\frac{d}{{dx}}(\tan x) - 7\frac{d}{{dx}}(\sec x)$
= 2 sec²x - 7sec x tan x

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MATHS 3 Marks Question

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