Question

Differentiate w.r.tx. 8^x/x^8.
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Answer 1

Answer:

Let y=8xx8⇒logy=log'8xx8.

⇒ddylogy.dydx=ddx[log8x−logx8]

⇒1y.dydx=ddx[x.log8−8logx]

On differenting w.r.t. x, we get

1y.dydx=log8.1−8.1x

⇒1y.dydx=log8−8x

∴dydx=y(log8−8x)=8xx8(log8−8x)

Answer 2

Answer:

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