Question

if y=e^x log x find dy/dx​

Answer 1

Answer:

we use first term e^x and 2nd logx

on differentiating use product rule,keep first term as it is and differentiate 2nd + differentiate 1st keep 2nd as it is

so,we get dy/dx is,

e^x/x+e^xlogx

Answer 2

$\tt \implies \large y = {e}^{x} \: logx$

Differentiating both sides :-

$\tt \implies \large \dfrac{dy}{dx} = \dfrac{d({e}^{x} \: logx)}{dx}$

Using product rule :-

$\boxed{\tt \dfrac{d(uv)}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx}}$

$\tt \implies \large \dfrac{dy}{dx} = logx \: \dfrac{d({e}^{x} \: )}{dx} +{e}^{x} \: \dfrac{d( \ log \: x)}{dx}$

$\tt \implies \large \dfrac{dy}{dx} = {e}^{x} \: logx +({e}^{x} \times \dfrac{1}{x} )$

$\tt \implies \large \dfrac{dy}{dx} = {e}^{x} \: logx + \dfrac{{e}^{x}}{x}$

Answer :

$\bf \: \large \dfrac{dy}{dx} = {e}^{x} \: logx + \dfrac{{e}^{x}}{x}$

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Learn more :

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x