Math, asked by stefern19, 5 months ago

if y = 2^x + cos3x then find dy/dx​

Answers

Answered by mathdude500
9

Question :-

\bf \:If \: y \:  =  {2}^{x}  + cos3x, \: find \: \dfrac{dy}{dx}

Answer

Given :-

\bf \:y \:  =  {2}^{x}  + cos3x

To Find :-

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

Formula used :

\bf \:\dfrac{d}{dx}  {a}^{x}  =  {a}^{x} loga

\bf \:\dfrac{d}{dx} cosx =  - sinx

\bf \:\dfrac{d}{dx} x = 1

Solution:-

\bf \:y \:  =  {2}^{x}  + cos3x

Differentiate w. r. t. x, we get

\bf\implies \:\dfrac{dy}{dx}  = \dfrac{d}{dx} ({2}^{x}  + cos3x)

\bf\implies \:\dfrac{dy}{dx}  = \dfrac{d}{dx}  {2}^{x}  +  \dfrac{d}{dx}cos3x

\bf\implies \:\dfrac{dy}{dx}  =  {2}^{x} log2 - sin3x\dfrac{d}{dx}3x

\bf\implies \:\dfrac{dy}{dx}  =  {2}^{x} log2 - sin3x \times (3)

\bf\implies \:\dfrac{dy}{dx}  =  {2}^{x} log2 - 3sin3x

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