# If y=8^(cos x), what is dy/dx?

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We have y = 8^cos x.

We use the chain rule to find the derivative of 8^(cos x)

Now the derivative of a^x is a^x log a and the derivative of cos x is -sin x.

y' = log 8 * 8^cos x * (cos x)'

=> y' = -log 8 * 8^cos x* sin x

**Therefore dy/dx = -log 8 * 8^cos x* sin x**

y = 8^cosx.

We take the logarithms of both sides:

lny = (cosx)(ln8).

We differentiate both sides with respect to x.

(1/y)(dy/dx) = (ln8)*(cosx)'.

dy/dx = y*(ln8)(-sinx).

dy/dx = (ln8)(8^(cosx))(-sinx)..

Therefore the differential coefficient of y = 8^(cosx) is dy/dx = -(ln8)(8^(cosx))sinx.

To determine the derivative of the given composed function, we'll differentiate both sides:

dy= [8^ (cos x)]'dx

We'll apply chain rule:

dy = 8^ (cos x)*(ln 8)*[(cos x)']dx

But (cos x)' = -sin x

dy = -8^ (cos x)*ln 8*(sin x) dx

We'll divide by dx both sides:

**dy/dx = -[8^ (cos x)]*(ln 8)*(sin x)**