Derivative of 7^(SinX)... something to do with log/ln?

You have to use the chain rule.

ln(7) * 7^(sin x) * cos (x)

let y = 7^(sinx)

take ln of both sides
ln y = ln (7^sinx)
ln y = sinx (ln7)
now use the rule for ln
y' /y = ln7(cosx)
y' = ln7(cosx) * (7^sinx)

To find the derivative of the function f(x) = 7^(sin(x)), you can use the chain rule and logarithmic differentiation.

Let's start with the chain rule. The chain rule states that if you have a composition of functions, then the derivative of the composition is equal to the derivative of the outer function multiplied by the derivative of the inner function.

In this case, the outer function is 7^x, and the inner function is sin(x). To find the derivative of the outer function, we can use logarithmic differentiation.

To apply logarithmic differentiation, take the natural logarithm (ln) of both sides of the equation f(x) = 7^(sin(x)), which gives us:

ln(f(x)) = ln(7^(sin(x)))

Next, use the logarithmic property that log(base a) (a^x) = x to rewrite the equation:

ln(f(x)) = sin(x) * ln(7)

Now, differentiate both sides of the equation with respect to x:

(d/dx) [ln(f(x))] = (d/dx) [sin(x) * ln(7)]

By applying the chain rule, we get:

[1/f(x)] * (d/dx) [f(x)] = cos(x) * ln(7)

Now, solve for (d/dx) [f(x)] which is the derivative of the original function f(x):

(d/dx) [f(x)] = f(x) * cos(x) * ln(7)

Substituting f(x) with 7^(sin(x)), we have:

(d/dx) [7^(sin(x))] = 7^(sin(x)) * cos(x) * ln(7)

So, the derivative of 7^(sin(x)) with respect to x is 7^(sin(x)) multiplied by cos(x) and ln(7).

Remember to always check your work and simplify the result if possible.