Question

derivative of y = 5^x + x^x​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = {5}^{x} + {x}^{x}$

Let assume that

$\rm :\longmapsto\:y = u + v$

where,

$\rm :\longmapsto\:u = {5}^{x}$

and

$\rm :\longmapsto\:v = {x}^{x}$

Now,

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

Consider

$\rm :\longmapsto\:u = {5}^{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}u = \dfrac{d}{dx}{5}^{x}$

$\rm :\longmapsto\:\dfrac{du}{dx}= {5}^{x} \: log5 - - - (2)$

Consider

$\rm :\longmapsto\:v = {x}^{x}$

On taking log on both sides, we get

$\rm :\longmapsto\:logv = log{x}^{x}$

$\rm :\longmapsto\:logv = x \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logv =\dfrac{d}{dx}[ x \: logx]$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = x\dfrac{d}{dx}logx \: + \: logx\dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = x \times \dfrac{1}{x} \: + \: logx \times 1$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = 1 + logx$

$\rm :\longmapsto\:\dfrac{dv}{dx} = v[1 + logx]$

$\rm :\longmapsto\:\dfrac{dv}{dx} = {x}^{x} [1 + logx] - - - - (3)$

On substituting the equation (2) and (3) in equation (1), we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {5}^{x}log5 + {x}^{x}(1 + logx)$

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Formula Used

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

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

$\boxed{\tt{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga \: }}$

$\boxed{\tt{ \dfrac{d}{dx}uv \: = \: v\dfrac{d}{dx}u \: + \: v\dfrac{d}{dx}u}}$

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More to Know

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

Answer 2

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

Given function is

$\rm :\longmapsto\:y = {5}^{x} + {x}^{x}$

Let assume that

$\rm :\longmapsto\:y = u + v$

where,

$\rm :\longmapsto\:u = {5}^{x}$

and

$\rm :\longmapsto\:v = {x}^{x}$

Now,

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

Consider

$\rm :\longmapsto\:u = {5}^{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}u = \dfrac{d}{dx}{5}^{x}$

$\rm :\longmapsto\:\dfrac{du}{dx}= {5}^{x} \: log5 - - - (2)$

Consider

$\rm :\longmapsto\:v = {x}^{x}$

On taking log on both sides, we get

$\rm :\longmapsto\:logv = log{x}^{x}$

$\rm :\longmapsto\:logv = x \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logv =\dfrac{d}{dx}[ x \: logx]$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = x\dfrac{d}{dx}logx \: + \: logx\dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = x \times \dfrac{1}{x} \: + \: logx \times 1$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = 1 + logx$

$\rm :\longmapsto\:\dfrac{dv}{dx} = v[1 + logx]$

$\rm :\longmapsto\:\dfrac{dv}{dx} = {x}^{x} [1 + logx] - - - - (3)$

On substituting the equation (2) and (3) in equation (1), we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {5}^{x}log5 + {x}^{x}(1 + logx)$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Formula Used

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

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

$\boxed{\tt{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga \: }}$

$\boxed{\tt{ \dfrac{d}{dx}uv \: = \: v\dfrac{d}{dx}u \: + \: v\dfrac{d}{dx}u}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

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