Differentiating given f(x)

**Starfoth** · October 21st 2012, 01:07 PM

Given y=f(x) with f(1)=4 and f'(1)=3, find:
A) g'(1) if g(x) = √(f(x))
B) h'(1) if h(x) = f(√(x))
And show how you got there so I can learn.

**Siron** · October 21st 2012, 01:11 PM

What have you tried? What is $g'(x)$ and $h'(x)$ ?

**johnsomeone** · October 21st 2012, 01:15 PM

These are both about using the chain rule.
I'll use a function other than the square root to give you an idea.

$\text{A) Suppose } h(x) = [f(x)]^3$

$\text{Then } h'(x) = 3[f(x)]^2f'(x)$

$\text{so } h'(1) = 3[f(1)]^2f'(1) = 3[4]^2(3) = (9)(16) = 144.$

$\text{B) Suppose } \alpha(x) = f(x^3)$

$\text{Then } \alpha'(x) = f'(x^3)(3x^2)$

$\text{so } \alpha'(1) = f'(1^3)(3(1^2)) = 3f'(1) = (3)(3) = 9.$

**Starfoth** · October 21st 2012, 01:17 PM

Originally Posted by johnsomeone

These are both about using the chain rule.
I'll use a function other than the square root to give you an idea.

$\text{A) Suppose } h(x) = [f(x)]^3$

$\text{Then } h'(x) = 3[f(x)]^2f'(x)$

$\text{so } h'(1) = 3[f(1)]^2f'(1) = 3[4]^2(3) = (9)(16) = 144.$

$\text{B) Suppose } \alpha(x) = f(x^3)$

$\text{Then } \alpha'(x) = f'(x^3)(3x^2)$

$\text{so } \alpha'(1) = f'(1^3)(3(1^2)) = 3f'(1) = (3)(3) = 9.$

Even better answer than I was expecting, now I can do it (and future problems) on my own, thank you.

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