Question

Find the differentiation by chain rule :
[sin (4x +9)]²​

Answer 1

Answer:

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Answer 2

Answer: 8 sin(4x+9).cos(4x+9)

Explanation:

$\sf We\:know\:that,\:chain\:rule\:states\:that,$

$\sf The\:derivative\:of\:x\:w.r.t\:v, can\:be\:given\:as:\\\\boxed{\sf f'(x)=\dfrac{dv}{du}.\dfrac{du}{dx} }$

$\sf [sin(4x+9)]^2=sin^2(4x+9)$

$\sf Differentiating\:w.r.t\:x:$

$\Rightarrow\sf \dfrac{d}{dx}sin^2(4x +9)$

$\Rightarrow\sf 2sin(4x +9)\dfrac{d}{dx}[sin(4x+9)] \quad - -(Using\:power\:rule)$

$\Rightarrow\sf 2sin(4x +9)\dfrac{d}{du}sin\:u\dfrac{d}{dx} (4x+9)$

$\Rightarrow\sf 2sin(4x +9).cos(4x+9)\dfrac{d}{dx} (4x+9)$

$\Rightarrow\sf 2sin(4x +9).cos(4x+9).4$

$\Rightarrow\boxed{\sf 8\:sin(4x +9).cos(4x+9)}$