# Thread: the integral of a^u where u is a function of x

1. ## the integral of a^u where u is a function of x

The integral of a^u where u is a function of x is given as

1/ln(a) a^u + C

but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0

so f'g=0 and for fg' we have

ua^(u-1)/ln(a) which does not look equal to a^u (but maybe it is).

I'm assuming u is a function of x but I'm setting it as u = x so that u' =1

taking the derivative of the integral should give back the original integral function correct? what am I doing wrong?

thanks!

2. ## Re: the integral of a^u where u is a function of x

Originally Posted by kingsolomonsgrave
The integral of a^u where u is a function of x is given as
1/ln(a) a^u + C

but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0
so f'g=0 and for fg' we have
You are missing the point. Forget the product rule.

If $\displaystyle a>0~\&~f(x)=a^x$ then $\displaystyle f'(x)=a^x\ln(a)$.

3. ## Re: the integral of a^u where u is a function of x

I see!

The integral of a^x ln(a) is a^x therefore the integral of a^x is a^x times 1/ln(a)

4. ## Re: the integral of a^u where u is a function of x

Originally Posted by kingsolomonsgrave
I see!

The integral of a^x ln(a) is a^x therefore the integral of a^x is a^x times 1/ln(a)
Note that this rule only holds for u = x. Any other function of x requires the chain rule.

-Dan

5. ## Re: the integral of a^u where u is a function of x

Originally Posted by kingsolomonsgrave
The integral of a^u where u is a function of x is given as

1/ln(a) a^u + C
The integral with respect to what variable? The integral of e^u with respect to u is (1/ln(a))a^u+ C:
$\displaystyle \int a^u du= \frac{1}{ln(a)} a^u+ C$

The integral with respect to x will depend upon exactly what function of x u is.

but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0
Why use the product rule? Here 1/ln(a) is a constant and the rule (Cf(x))'= Cf'(x) is much simpler.

so f'g=0 and for fg' we have

ua^(u-1)/ln(a) which does not look equal to a^u (but maybe it is).
Wrong rule. The derivative of $\displaystyle u^a$, with respect to u, is $\displaystyle au^{a- 1}$ but that only applies when the variable is in the base and the exponent is constant. The derivative of $\displaystyle a^x$ is $\displaystyle ln(a) a^x$.

I'm assuming u is a function of x but I'm setting it as u = x so that u' =1

taking the derivative of the integral should give back the original integral function correct? what am I doing wrong?

thanks!

6. ## Re: the integral of a^u where u is a function of x

Originally Posted by kingsolomonsgrave
The integral of a^u where u is a function of x is given as

1/ln(a) a^u + C

but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0

so f'g=0 and for fg' we have

ua^(u-1)/ln(a) which does not look equal to a^u (but maybe it is).

I'm assuming u is a function of x but I'm setting it as u = x so that u' =1

taking the derivative of the integral should give back the original integral function correct? what am I doing wrong?

thanks!
\displaystyle \displaystyle \begin{align*} \int{a^x\,dx} &= \int{ e^{\ln{ \left( a^x \right) } } \, dx} \\ &= \int{ e^{x\ln{(a)}}\,dx} \\ &= \frac{1}{\ln{(a)}}\,e^{x\ln{(a)}} + C \\ &= \frac{1}{\ln{(a)}} \, e^{\ln{ \left( a^x \right) } } + C \\ &= \frac{1}{\ln{(a)}}\,a^x + C \end{align*}