differentiate by using the chain rule

Question

y=(2׳−7)^4

Answer 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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Answer 2

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'

Answer 3

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)

= 24x^2(2x^3-7) <==Answer