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# differentiate by using the chain rule

y=(2×³−7)^4

Just a question about your problem, I just wanted to make sure if the problem is not a typo. Is the 2x part is written in the form on the exponential function such as 2xor do you mean 2x3? (where only the 3 is the exponent).
no is written like this (2x^2-7)^4

### 3 Answers by Expert Tutors

William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
1
[(2x)2-7)]4

Rewrite as (-7 + 4x2)4

Let u = 4x2 - 7

[d/dx] [(4x2 - 7)]4 = [du4/du]*[du/dx] where u = 4x2 - 7 and [d/du](u4) = 4u3

so that

[d/dx] [(4x2 - 7)]4 = 4(-7 + 4x2)3{[d/dx](-7 + 4x2)}

[d/dx] [(4x2 - 7)]4 = 4(-7 + 4x2)3(0 + 8x) = (8x)(4)(-7 + 4x2)3

[d/dx] [(4x2 - 7)]4 = (32x)(-7 + 4x2)3
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Stephanie B. | Math, Science and Language TutorMath, Science and Language Tutor
4.8 4.8 (99 lesson ratings) (99)
0
Chain rule:

(2x^2-7)^4 = f (as per you stating this is how it should be written)

Follow the steps for power rule, but now you'll have to differentiate what is inside the parenthesis. Technically you have done this with say x^2, which is 2x * 1 (derivative of x is 1)

4*(2x^2-7)^3 * (4x)

16x * (2x^2-7)^3 = f'

Original Problem: f = (2x^3 - 7)^4

Same thing as we did before.

4 * (2x^3 - 7)^3 * (6x^2)

= (24x^2)*(2x^3 - 7)^3 = f'
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
0
Assume you mean y = (2x^3-7)^4
Using chain rule,
y'
= 4(2x^3-7)^3 (2x^3-7)'
= 4(2x^3-7)^3 (6x^2)