# If y= 3^ (sin x), what is dy/dx?

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### 3 Answers

It is given that y = 3^ (sin x)

We take the natural logarithm of both the sides and use the relation ln a^b = b*ln a.

=> ln y = ln (3^sin x)

=> ln y = (sin x) (ln 3)

Now differentiate both the sides with respect to x

=> (1/y) (dy/dx) = (ln 3) cos x

=> dy/dx = (ln 3)*y*cos x

=> dy /dx = 3^ (sin x)*(cos x)* ln 3

**Therefore if y= 3^ (sin x), dy /dx = 3^ (sin x)*(cos x)* ln 3.**

y = 3^(sinx).

To find dy/dx.

We know that y = a^f(x) has the differential coefficient , dy/dx = {a^f(x)}' = {(lna)a^f(x)} f'(x).

Therefore a = 3, f(x) = sinx , f'(x) = (sinx)' = cosx..

Therefore dy/dx = {3^(sinx)}'

dy/dx = {(ln3) 3^(sinx) }(sinx)'

dy/dx = {(ln3)3^(sinx)} cosx.

Therefore the differential coefficient of 3^(sinx) is (ln3^(sinx))(cosx).

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

dy= [3^ (sin x)]'dx

We'll apply chain rule:

dy = 3^ (sin x)*ln 3*(sin x)'dx

But (sin x)' = cos x

dy = 3^ (sin x)*ln 3*(cos x) dx

We'll divide by dx both sides:

**dy/dx = (ln 3)*[3^ (sin x)]*(cos x)**