Question

Help with finding the first derivative?

Answer 1

The derivative of a^x is a^xln(a) isn't it?

Answer 2

This is what I have so far, but after this I'm lost.

Answer 3

This is what I have. Possibly definitely wrong.

the equation button isn't working so I can't explain atm

Answer 4

You need to treat that derivative as a product - (f*g)' = f'*g+f*g', where f is the 2^4x+ln2 and g is x^2.

The issue is that when you applied the derivative to the 2^4x+ln2 (the so called "f") you forgot to apply the derivative to ln2. Remember, (a+b)' is a' + b, ergo (2^4x+ln2)' * x^2 is ( (2^4x)' + (ln2)' ) *x^2. Since ln2 is a constant, (ln2)'=0.

Answer 5

* (a+b)' is a' + b', sorry

Answer 6

Ohh okay then. I've attached what I've done now, but what can I factor now (besides x)?

Answer 7

Yup, looks good to me.