Differential using chain Rule

**charikaar** · April 10th 2010, 01:34 PM

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.

**maddas** · 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?

**charikaar** · April 10th 2010, 01:55 PM

sorry still can't solve it with exponential!

**maddas** · 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.

You can do eet!