# Exponential functions with integral

1. Jan 2, 2010

### b00tofuu

1. The problem statement, all variables and given/known data
problem 1
diff([f(tanx)], x) = x^2; prove that diff[f(u)]=(tan^-1(u))^2/(1+u^2)
problem 2
int(f(u), u = 0 .. 1) = f(t)-1; prove that f(a+b)=f(a)f(b) for every a,b ∈R

3. The attempt at a solution
problem 1
i don't know how to answer this question... taking the integral of f'(tan x) =x^3/3+C doesn't seem to help....

problem 2
the function seems to be the exp function where it's itself its derivative and e^0=1. but is it correct to state that the function is the exponential function directly?

btw,i'm right now learning the technique of integration.

2. Jan 2, 2010

### HallsofIvy

Re: integral

To find df(tan x)/dx, use the chain rule.
Taking u= tan(x), df(u)/dx= df/du (du/dx) and here du/dx= d(tan x)dx= sec^2(x) so df(tan(x))dx= df(u)/du(sec^2(x))= x^2. From that df(u)/du= x^2/sec^2(x). Now use the fact that u= tan(x) so x= tan^(-1)(u). What is sec^2(x) in terms of tan(x)?

$\int_0^1 f(u)du= f(t)- 1$ is impossible. The left side is a constant, not a function of t. Did you mean $\int_0^t f(u)du= f(t)- 1$? If so, differentiating both sides with respect to t gives f(t)= f'(t). What function satifies that?

3. Jan 2, 2010

### b00tofuu

Re: integral

thank u... i understand problem 1 now...

for problem 2, i'm sorry. you're right, it's from 0 to t not 1... thank u...