Question

differentiate w.r.t x log(tan^3x .sin^4x(x^2+7)^7)

Answer 1

$\textbf{Given:}$

$y=\log(tan^3x\,sin^4x\,(x^2+7)^7)$

$\textbf{To find:}$

$\text{Derivative of y}$

$\text{Consider,}$

$y=\log(tan^3x\,sin^4x\,(x^2+7)^7)$

$\text{Using product rule of logarithm}$

$y=\log(tan^3x)+\log(sin^4x)+\log(x^2+7)^7$

$y=\log(tanx)^3+\log(sinx)^4+\log(x^2+7)^7$

$\text{Using power rule of logarithm}$

$y=3\,\log(tanx)+4\,\log(sinx)+7\,\log(x^2+7)$

$\text{Differentiate with respect to x}$

$\dfrac{dy}{dx}=3(\dfrac{1}{tanx})sec^2x+4(\dfrac{1}{sinx})cosx+7\,(\dfrac{1}{x^2+7})2x$

$\dfrac{dy}{dx}=3(\dfrac{sec^2x}{tanx})+4\,cotx+7\,(\dfrac{2x}{x^2+7})$

$\dfrac{dy}{dx}=3(\dfrac{1}{sinx\,cosx})+4\,cotx+7\,(\dfrac{2x}{x^2+7})$

$\implies\bf\dfrac{dy}{dx}=3\,cosecx\,secx+4\,cotx+\dfrac{14x}{x^2+7}$

$\therefore\textbf{The derivative of \bf\log(tan^3x\,sin^4x\,(x^2+7)^7) is \bf\,3\,cosecx\,secx+4\,cotx+\dfrac{14x}{x^2+7} }$

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