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  • Differentiate w.r.t. x the function in Exercises 1 to 11. 5 x ^ 3 cos 2x

Q3  Differentiate w.r.t. x the function in Exercises 1 to 11.

        ( 5 x) ^{ 3 \cos 2x }

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Given function is
y=( 5 x) ^{ 3 \cos 2x }
Take, log on both the sides
\log y = 3\cos 2x\log 5x
Now, differentiation w.r.t. x is
By using product rule
\frac{1}{y}.\frac{dy}{dx} = 3.(-2\sin 2x)\log 5x + 3\cos 2x.\frac{1}{5x}.5= -6\sin2x\log 5x +\frac{3\cos 2x}{x}\\ \frac{dy}{dx} = y.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )\\ \frac{dy}{dx} = (5x)^{3\cos 2x}.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )

Therefore, differentiation w.r.t. x is   (5x)^{3\cos 2x}.\left ( \frac{3\cos 2x}{x}-6\sin2x\log 5x \right )

Posted by

Gautam harsolia

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