# Differential using chain Rule

• April 10th 2010, 01:34 PM
charikaar
Differential using chain Rule
suppose that f:R-->R is continuous. using the chain rule, or otherwise, show that the function F defined by
$F(x) = \exp(\int_0^xf)$ is differentiable and find its derivative.

Fundamental Theorem of calculus say that the derivative of $\int_a^xf(t)dt = f(x)$ I tried to use this to solve the above but no vain.
thanks for any help.
• April 10th 2010, 01:45 PM
You have the FTC wrong: it is $\int_a^b f'(x)\, \mathrm{d}x = f(b) - f(a)$, or equivalently, $\frac{\mathrm{d}}{\mathrm{d}x}\int_a^xf(s)\,\mathr m{d}s = f(x)$.

Does this help?
• April 10th 2010, 01:55 PM
charikaar
sorry still can't solve it with exponential!
• April 10th 2010, 02:02 PM
Let $h(x) = \int_0^x f$. Then $F(x) = \exp(h(x))$. Applying the chain rule gives $F'(x) = \exp(h(x))\cdot h'(x)$. The exponential is now out of the way and you only have to differentiate h. Use the FTC for this.
Define the function $g(t)$ to be the antiderivative of $f(t)$
Then, $\int_{0}^{x} f(t) dt = g(x) - g(0)$
So, $F(x) = e^{g(x) - g(0)}$