Question

Differentiate the following w.r.t x​

Answer 1

Question :-

Differentiate the following function with respect to x :-

$\rm \: {e}^{x}( {sec}^{2}x + tanx) + \dfrac{cosx}{x} - sinxlogx - {5}^{x}\bigg(\dfrac{2xlog5 + 1}{2 \sqrt{x} } \bigg)$

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

Given function is

$\rm \: {e}^{x}( {sec}^{2}x + tanx) + \dfrac{cosx}{x} - sinxlogx - {5}^{x}\bigg(\dfrac{2xlog5 + 1}{2 \sqrt{x} } \bigg)$

Let assume that

$\rm \: y = u + v - w - z$

So, that

$\rm \: \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} - \dfrac{dw}{dx} - \dfrac{dz}{dx} - - - (1)$

where,

$\rm \: u = {e}^{x}( {sec}^{2}x + tanx) \\ \rm \: v = \dfrac{cosx}{x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: w = sinxlogx \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: z = {5}^{x}\bigg(\dfrac{2xlog5 + 1}{2 \sqrt{x} } \bigg)$

Now, Consider

$\rm \: u = {e}^{x}( {sec}^{2}x + tanx)$

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

$\rm \: \dfrac{du}{dx} = {e}^{x}\dfrac{d}{dx}( {sec}^{2}x + tanx) + ( {sec}^{2}x + tanx)\dfrac{d}{dx}{e}^{x}$

$\rm \: = {e}^{x}(2secx \times secxtanx + {sec}^{2}x) + {e}^{x}( {sec}^{2}x + tanx)$

$\rm \: = {e}^{x}(2sec^{2} x tanx + {sec}^{2}x) + {e}^{x}( {sec}^{2}x + tanx)$

$\rm \: = {e}^{x}(2sec^{2} x tanx + {sec}^{2}x+ {sec}^{2}x + tanx)$

$\rm \: = {e}^{x}(2sec^{2} x tanx + 2{sec}^{2}x + tanx)$

$\rm\implies \:\dfrac{du}{dx} = {e}^{x}(2sec^{2} x tanx + 2{sec}^{2}x + tanx) - - (2)$

Now, Consider

$\rm \: v = \dfrac{cosx}{x}$

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

$\rm \: \dfrac{dv}{dx} = \frac{x\dfrac{d}{dx}cosx - cosx\dfrac{d}{dx}x}{ {x}^{2} }$

$\rm \: \dfrac{dv}{dx} = \frac{ - xsinx - cosx}{ {x}^{2} }$

So,

$\rm\implies \:\rm \: \dfrac{dv}{dx} = \frac{ - xsinx - cosx}{ {x}^{2} } - - - (3)$

Now, Consider

$\rm \: w = sinx \: logx$

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

$\rm \: \dfrac{dw}{dx} = sinx \: \dfrac{d}{dx}logx + logx\dfrac{d}{dx}sinx$

$\rm \: \dfrac{dw}{dx} = sinx \times \frac{1}{x} + logx \: cosx$

$\rm\implies \:\rm \: \dfrac{dw}{dx} = \frac{sinx}{x} + logx \: cosx - - - (4)$

Now, Consider

$\rm \: z = {5}^{x}\bigg(\dfrac{2xlog5 + 1}{2 \sqrt{x} } \bigg)$

$\rm \: z = {5}^{x}\bigg(log5 \sqrt{x} + \frac{1}{2 \sqrt{x} } \bigg)$

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

$\rm \: \dfrac{dz}{dx} = {5}^{x}\dfrac{d}{dx}\bigg(log5 \sqrt{x} + \frac{1}{2 \sqrt{x} } \bigg) + \bigg(log5 \sqrt{x} + \frac{1}{2 \sqrt{x} } \bigg)\dfrac{d}{dx} {5}^{x}$

$\rm \: = {5}^{x}\bigg(\dfrac{log5}{2 \sqrt{x} } - \dfrac{1}{4x \sqrt{x} } \bigg) + \bigg(log5 \sqrt{x} + \dfrac{1}{2 \sqrt{x} } \bigg) {5}^{x}log5$

$\rm\implies \:\dfrac{dz}{dx} = {5}^{x}\bigg(\dfrac{log5}{2 \sqrt{x} } - \dfrac{1}{4x \sqrt{x} } \bigg) + \bigg(log5 \sqrt{x} + \dfrac{1}{2 \sqrt{x} } \bigg) {5}^{x}log5 - - (5)$

On substituting the values from equation (2), (3), (4), (5) in equation (1), we get

$\bf\implies \:\dfrac{dy}{dx}$

$\rm \: = {e}^{x}(2sec^{2} x tanx + 2{sec}^{2}x + tanx) + \\ \rm \: \frac{ - xsinx - cosx}{ {x}^{2} } - \frac{sinx}{x} - logx \: cosx - \\ \rm \: {5}^{x}\bigg(\dfrac{log5}{2 \sqrt{x} } - \dfrac{1}{4x \sqrt{x} } \bigg) - \bigg(log5 \sqrt{x} + \dfrac{1}{2 \sqrt{x} } \bigg) {5}^{x}log5$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {x}^{n}\\ \\ \sf {nx}^{n - 1} & \sf {e}^{x} \end{array}} \\ \end{gathered}$