can you help me with the math problem please?

Question

i need help finding the derivative of this- 3rd root of ((7x-12)^5)

i know we can change it to (7x-12)^5/3 but i dont know where to go from there. thanks, anyhelp is appreciated.

Answer 1

This I can explain with chain rule
let function be f= (7x-12)^5/3
let t= (7x-12)
using the formala d/dt(t^n) = nt^(n-1)

now d/dt(t^5/3) = (5/3) t^(5/3-1) = (5/3) t^2/3

now dt/dx = d/dx(7x-12) = 7
using chain rule

df/dx = df/dt * dt/dx
= (5/3)(t)^(2/3) * 7
= 35/3 (7x-12)^(2/3)

Answer 2

Note that the 3rd root of (7x - 12)^5 can be expressed as (7x - 12)^(5/3). In order to differentiate this, we apply the chain rule that is:

Dx { (7x - 12)^(5/3) } = (5/3) * (7x - 12) ^(5/3 - 1) * Dx {7x - 12}
= (5/3) * (7x - 12)^(2/3) * Dx {7x - 12}
= (5/3) * (7x - 12)^(2/3) * [ Dx {7x} - Dx {12} ]
= (5/3) * (7x - 12)^(2/3) * [ 7 Dx{x} - 0 ]
= (5/3) * (7x - 12)^(2/3) * 7(1)
= (5/3) * (7x - 12)^(2/3) * 7
= (35/3) * (7x - 12)^(2/3).

Therefore, the derivative of (7x - 12)^(5/3) is (35/3) * (7x - 12)^(2/3).

Answer 3

d/dx (7x - 12)^(5/3)
= (7) (5/3) (7x - 12)^(2/3)
= (35/3) (7x - 12)^(2/3)

The derivative of a function raised to a power is the power, times the function to the one lower power, times the derivative of the function.

Answer 4

then u use derivative of x^p is px^(p-1)

(7x-12)^5/3

{ (5/3) (7x-12)^2/3 } * 7

times 7 is the chain rule f((7x)) : f'(7x) * 7

Answer 5

yes. U can change it to (7x-12)^5/3.

Now,lets assume (7x-12)^5/3 = U;
So what u need to find is dU/dx.

Take log on both sides , u get

log U = 5/3 * log(7x-12)

on differentiating wrt x,
1/U dU/dx = 5/3* 1 / (7x-12) * 7 = 35 / (21x-36)

hence, dU/dx= (7x-12)^5/3 * 35 / (21x-36) :-)

Answer 6

y = (7x-12)^5/3
y ' = 5/3 * (7x-12)^(5/3 - 1) * 7
{7 is the deravate from 7x - 12}

Th