Compute f'(x) in three different ways

Question

f(x)=(2x^2-5)^2

1.) multiplying out and then differentiating

2.) Using the product rule

3.) Using the chain rule

Show that the results coincide

Answer 1

f(x)=(2x^2 - 5)^2

A) Multiply it out:

f(x) = 4x^4 - 20x^2 + 25

f'(x) = 16x^3 - 40x

B) Product Rule:

d(u*v) = udv + vdu

in this case, u and v are both 2x^2 - 5

So du and dv are both 4x

So:

(2x^2 - 5) * 4x + (2x^2 - 5) * 4x

(2x^2 - 5) * 8x

16x^3 - 40x

C) Chain Rule:

If f(x) =g(h(x)), then f'(x) = g'(h(x)) * h'(x)

So f'(x) = 2*(2x^2 - 5)*(4x)

16x^3 - 40x

Answer 2

Multiplying out the differentiate:

f(x)=(2x²-5)(2x²-5) = 4x⁴-20x²+25

f'(x) = 16x³-40x

Product rule:

f'(x) = (2x²-5)(4x)+(2x²-5)(4x) Distribute

f'(x) = 8x³-20x+8x³-20x Combine like terms

f'(x) = 16x³-40x

Chain Rule:

f'(x) = 2(2x²-5)(4x) Multiply

f'(x) = 8x(2x²-5) Distribute

f'(x) = 16x³-40x

As expected, all 3 results are the same.