Question

Important Question for jee mains.
mathamatics :- Practice set 1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f'(x) = tanx$

$\rm :\longmapsto\:g'(x) = x$

$\rm :\longmapsto\:y = f( {x}^{3})$

and

$\rm :\longmapsto\:z = g( {x}^{5})$

Now,

Consider,

$\rm :\longmapsto\:y = f( {x}^{3})$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} f( {x}^{3})$

$\rm :\longmapsto\:\dfrac{dy}{dx} = f'( {x}^{3})\dfrac{d}{dx} {x}^{3}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = f'( {x}^{3}) \times {3x}^{2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = tan{x}^{3}\times {3x}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm \because \: \:f'(x) = tanx}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {3x}^{2} tan{x}^{3} - - - (1)$

Now,

Consider,

$\rm :\longmapsto\:z = g( {x}^{5})$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}z =\dfrac{d}{dx} g( {x}^{5})$

$\rm :\longmapsto\:\dfrac{dz}{dx} =g'( {x}^{5}) \dfrac{d}{dx} {x}^{5}$

$\rm :\longmapsto\:\dfrac{dz}{dx} =g'( {x}^{5}) \times 5 {x}^{4}$

$\rm :\longmapsto\:\dfrac{dz}{dx} = {5x}^{4} \times {x}^{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm \because \: \:g'(x) = x}$

$\rm :\longmapsto\:\dfrac{dz}{dx} = {5x}^{9} - - - (2)$

Now, Divide equation (1) by (2), we get

$\rm :\longmapsto\:\dfrac{dy}{dz} = \dfrac{ {3x}^{2} \: tan {x}^{3} }{5 {x}^{9} }$

$\rm :\longmapsto\:\dfrac{dy}{dz} = \dfrac{3\: tan {x}^{3} }{5 {x}^{7} }$

Additional Information ;-

$\boxed{ \sf{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {x} = 1}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {k} = 0}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {sinx} = cosx}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {cosx} = - sinx}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {cosecx} = - cosecx \: cotx}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {secx} = secx \: tanx}}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {tanx} = sec^{2} x }}$

$\boxed{ \sf{ \:\dfrac{d}{dx} {cotx} = - cosec^{2} x }}$

Answer 2

$\huge\red{A}\pink{N}\orange{S}\green{W}\blue{E}\gray{R}</p><p>$

$\small\red{y = f(x) ^3, z = g(x^5)}$

$\small\pink{ \frac{dy}{dx} = f'(x^3) × 3x^2 and \: \frac{dz}{dx} = g'(x {}^{5} \times 5x {}^{4} )}$

$\small\orange{ \frac{dy}{dz} = \frac{ f' (x {}^{3}) }{g'(x {}^{5}) } \times \frac{3}{5} \frac{1}{x {}^{2} } }$

$\small\green{ \frac{dy}{dz} = \frac{3}{5x {}^{2} } \times \frac{f'(x {}^{3}) }{g'(x {}^{5}) } }$

$\huge\blue{ Answer = \frac{3}{x {}^{2} } \: \frac{tanx {}^{3} }{sec {}^{5} } }$