Question

derivative of y=5^lnsinx​

Answer 1

Answer:

$dy = ln5( {5}^{lnsinx} )(cotx)dx$

Step-by-step explanation:

$y = {5}^{lnsinx}$

Taking natural log on both sides.

$lny = ln( {5}^{lnsinx} )$

$lny = ln5(lnsinx )$

Differentiate with respect to x

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{sinx} cosx(ln5)$

$\frac{dy}{dx} = y(cotx)ln5$

$\frac{dy}{dx} = ln5( {5}^{lnsinx} )(cotx)$

Thank you.

Answer 2

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