Question

what derivative of this function??

Answer 1

let u =4x^2-4x and solve

\[\frac{d3^u}{du}\frac{du}{dx}\]

Answer 2

\[=ln(3)3^u(\frac{d(4x^2-4x)}{dx})\]

Answer 3

2x^4-4x not 4x^2-4x

Answer 4

\[=ln(3)3^{2x^4-4x}(8x^3-4)\]

Answer 5

Use chain rule it is 3^(2x^4-4x) (8x^3-4) ln3

Answer 6

If you have f(g(h(x))) differerential of this is f'(g(h(x)) g'(h(x)) h'(x)

Answer 7

why 8x^3-4 not 8x^-4 ??

Answer 8

Differential of 2x^4-4x=8x^3 -4 (differential of x^n is n (x^(n-1))

Answer 9

In this case if we take 3^x as f(x) and 2x^4-4x as g(x) then the net function is f(g(x)) hence by chain rule differential is f'(g(x))*g'(x) here g'(x) is 8x^3 -4

Answer 10

oh,i am sorry not carefully....