Question

If y = 6x² + 4x + log x, then d²y/dx² will be *

Answer 1

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

$\rm :\longmapsto\:y = {6x}^{2} + 4x + logx$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} }$

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

Given that

$\rm :\longmapsto\:y = {6x}^{2} + 4x + logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} ({6x}^{2} + 4x + logx)$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx} (u + v) = \dfrac{d}{dx} u + \dfrac{d}{dx} v}}$

and

$\boxed{ \bf{ \: \dfrac{d}{dx} k \: f(x) \: = \: k \: \dfrac{d}{dx} f(x) \: }}$

Using these result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =6\dfrac{d}{dx}{x}^{2} + 4\dfrac{d}{dx} x + \dfrac{d}{dx} logx$

Now, we know that

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

and

$\boxed{ \bf{ \: \dfrac{d}{dx} logx = \frac{1}{x}}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 6(2x) + 4(1) + \dfrac{1}{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 12x + 4 + \dfrac{1}{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} \dfrac{dy}{dx} = \dfrac{d}{dx} \bigg(12x + 4 + \dfrac{1}{x} \bigg)$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} } = 12\dfrac{d}{dx} x + \dfrac{d}{dx} 4 + \dfrac{d}{dx} {x}^{ - 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} } = 12(1) + 0 + ( - 1) {x}^{ - 1 - 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} } = 12- {x}^{ - 2}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{ {dx}^{2} } = 12- \dfrac{1}{ {x}^{2} }$

Additional Information :-

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

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

$\boxed{ \bf{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} logx = \frac{1}{x}}}$

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

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

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

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

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

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