Question

find d
$\frac{d {}^{2}y }{dx {}^{2} } \: \:if \: \: y = e {}^{5x}$
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Answer 1

Given :

$y = {e}^{5x}$

To find :

$\dfrac{d ^{2}y }{dx ^{2} }$

Solution :

$\implies \: y = {e}^{5x}$

$\implies \: \dfrac{dy}{dx} = \dfrac{d({e}^{5x})}{dx}$

we know that :-

• d(e^t)/dt = e^t

Using chain rule :-

$\implies \: \dfrac{dy}{dx} = {e}^{5x} \: \dfrac{d({5x})}{dx}$

$\implies \: \dfrac{dy}{dx} =5 {e}^{5x}$

Now :-

$\implies \dfrac{d ^{2}y }{dx ^{2} } = \dfrac{d(5 {e}^{5x})}{dx}$

$\implies \dfrac{d ^{2}y }{dx ^{2} } =5{e}^{5x} \times \dfrac{d({5x})}{dx}$

$\implies \dfrac{d ^{2}y }{dx ^{2} } =25 \: {e}^{5x}$

Answer :

$\dfrac{d ^{2}y }{dx ^{2} } =25 \: {e}^{5x}$

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Learn more :-

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

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

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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Answer 2

Given ,

The function is $\tt y = e {}^{5x}$

Differentiaing y wrt to x , we get

$\tt \frac{dy}{dx} = \frac{d}{dx} {e}^{5x}$

$\tt \frac{dy}{dx} = {e}^{5x} \frac{d}{dx} 5x$

$\tt \frac{dy}{dx} = {e}^{5x} \times 5$

$\tt \frac{dy}{dx} = 5{e}^{5x}$

Again differentiaing wrt to x , we get

$\tt \frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} 5{e}^{5x}$

$\tt \frac{ {d}^{2} y}{d {x}^{2} } = 5{e}^{5x}\frac{d}{dx} 5x$

$\tt \frac{ {d}^{2} y}{d {x}^{2} } = 5{e}^{5x} \times 5$

$\tt \frac{ {d}^{2} y}{d {x}^{2} } = 25{e}^{5x}$

Remmember :

$\tt \frac{d}{dx} {e}^{x}= {e}^{x}$

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