Question

y= cot x. 5^x differentiate w.r.t x​

Answer 1

GIVEN :–

$\\\implies \bf y= \cot(x).5^x$

TO FIND :–

• Differentiation = ?

SOLUTION :–

$\\\implies \bf y= \cot(x).5^x$

• Differentiate with respect to 'x' –

• We know that –

$\\\longrightarrow\red{ \bf \dfrac{d(u.v)}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx} }$

• So that –

$\\\implies \bf \dfrac{dy}{dx}= \cot(x) \dfrac{d(5^x)}{dx} +5^x \dfrac{d \{\cot(x)\}}{dx}$

• We also know that –

$\\\longrightarrow \red{\bf \dfrac{d(a^x)}{dx} = {a}^{x} log(x)}$

$\\\longrightarrow \red{\bf \dfrac{d( \cot x)}{dx} = - cosec^{2}x}$

• So that –

$\\\implies \bf \dfrac{dy}{dx}= \cot(x) 5^x log(x) +5^x (-cosec^{2}x)$

$\\\implies{\green{ \large{ \boxed{ \bf\dfrac{dy}{dx} =5^x[\cot(x)log(x) -cosec^{2}x]}}}}$

Answer 2

Answer:

$y = \cot \: {x}^{3} \\ put \: {x}^{3} = t \: \: and \: \: 3 {x}^{2} d x = dt \\ y = \cot \: t \\ diffferentiation \: with \: respect \: to \: t \\ dy = \: {cosec}^{2} tdt \\ put \: and \: value \: so \\ dy = - {3x}^{2} {cosec}^{2} {x}^{3} dx \\ \frac{dy}{dx} = {3x}^{2} {cosec}^{2} {x}^{3}$

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