Find the derivative of g(x)=cos(10^(2x))?

g(x) = cos(10^(2x))

First of all, I personally like to reduce the number of functions I use, because the chain rule can become a nightmare if too many functions are involved. Here, we have 3 functions: cos, 10^, and 2x.

But 10^(2x) can be changed to (10^2)^x, or simply 100^x.

g(x) = cos(100^x)

Differentiate. We need the derivatives of (1) cos(x), and (2) 100^x.

The derivative of cos(x) is -sin(x), and the derivative of 100^x is 100^x ln(100).

Apply the chain rule to get

g'(x) = -sin(100^x) (100^x) ln(100)

Let's rewrite this neater.

g'(x) = -(100^x) ln(100) sin(100^x)

The correct answer is g'(x) = -2 * ln(10) * sin(100^x) * 100^x

Use the chain rule.

-2*ln(10)*10^(2*x)*sin(10^(2*x))

This is a crazy function and will never appear in engineering.

Find the derivative of g(x)=cos(10^(2x))?

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Let u = 10^2x, then ln u = ln 10^2x = 2x ln10 = (ln 10).2x

u ' (x) / u = ln 10 . (2) = 2 ln 10 = ln 10^2

u ' (x) = (ln 10^2) u = (ln 10^2). 10^2x

Hence, g ' (x) = (- sin 10^2x ) (ln 10^2 ).(10^2x)

= - (sin 10^2x) (ln 10^2 ).(10^2x)

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Hope this helps!