# Calculus differentiation

• Mar 23rd 2008, 07:26 PM
Gorie
Calculus differentiation
f(x) = 3 + the integral from 8 to x cubed of f(cubed root of t) dt
• Mar 23rd 2008, 07:33 PM
mr fantastic
Quote:

Originally Posted by Gorie
f(x) = 3 + the integral from 8 to x cubed of f(cubed root of t) dt

$\frac{d}{dx} \left[ \int_{8}^{x^3} f(t^{1/3}) \, dt\right]$

Let $u = x^3$ and use the Fundamental Theorem of Calculus and the chain rule:

$= \frac{d}{du} \left[ \int_{8}^{u} f(t^{1/3}) \, dt\right] \, \times \, \frac{du}{dx}$

$= f(u^{1/3}) \, \times \, 3x^2$

$= f(x) \, \times \, 3x^2 = 3x^2 \, f(x)$.
• Mar 23rd 2008, 07:37 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
$\frac{d}{dx} \left[ \int_{8}^{x^3} f(t^{1/3}) \, dt\right]$

Let $u = x^3$ and use the Fundamental Theorem of Calculus and the chain rule:

$= \frac{d}{du} \left[ \int_{8}^{u} f(t^{1/3}) \, dt\right] \, \times \, \frac{du}{dx}$

$= f(u^{1/3}) \, \times \, 3x^2$

$= f(x) \, \times \, 3x^2 = 3x^2 \, f(x)$.

$\frac{d}{dx} \left[ \int_{8}^{x^3} f(t^{1/3}) \, dt\right]$

Alternatively, make the substitution $t^{1/3} = u \Rightarrow t = u^3$. Then you have

$\frac{d}{dx} \left[ \int_{8}^{x} f(u) \, 3u^2 \, du\right]$

Now apply the Fundamental Theorem of Calculus:

$= f(x) \, 3x^2$.