# Thread: Finding Deriv when # is raised ^x

1. ## Finding Deriv when # is raised ^x

This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

First simply finding d/dx(4^x)?

The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.

This is all part of a larger problem which is:

Find f'(x): (4^x*x^(1/3))/tanx

Do I use the product rule seperately to get the deriv of the numerator and then do the quotient rule for the new numerator/tan(x)?

2. Let $\displaystyle 4^x = e^{ln(4^x)} = e^{xln(4)}$. Thus the derivative, using the chain rule equals: $\displaystyle e^{xln(4)}*ln(4)$. From the first equality, we know this equals $\displaystyle 4^x*ln(4)$. The TI-89 says $\displaystyle 4^x*2ln(2)$ because $\displaystyle ln(4)=ln(2^2)=2ln(2)$.

In general, $\displaystyle \frac{d}{\,dx}c^x = ln(c)*c^x$, where $\displaystyle c$ is a constant.

3. Originally Posted by mattc
This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

First simply finding d/dx(4^x)?

The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.

This is all part of a larger problem which is:

Find f'(x): (4^x*x^(1/3))/tanx

Do I use the product rule seperately to get the deriv of the numerator and then do the quotient rule for the new numerator/tan(x)?
Let y = 4^x so that $\displaystyle ln(y) = x\ln{4} = 2x\ln{(2)}$

Differentiating:
$\displaystyle (\frac{1}{y})(\frac{dy}{dx}) = 2\ln{(2)}$

multiply by y:

$\displaystyle \frac{dy}{dx} = 2\ln{(2)} y$

Recall that y= 4^x so we get

$\displaystyle \frac{dy}{dx} = 2\ln{(2)}*4^x$

In general: $\displaystyle \frac{d}{dx}(a^x) = a\ln{(a^x)}$

4. Originally Posted by mattc
This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

First simply finding d/dx(4^x)?

The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.
The rule is

$\displaystyle \frac d{dx}\left[a^x\right]=(\ln a)a^x.$

To see why this is so, you may rewrite $\displaystyle a^x$ as $\displaystyle e^{x\ln a}$ and apply the more familiar rule for differentiating $\displaystyle e^x.$

5. $\displaystyle y = a^x$

$\displaystyle \ln{y} = \ln{a^x}$

$\displaystyle \ln{y} = x \cdot \ln{a}$

$\displaystyle \frac{d}{dx}[\ln{y} = x \cdot \ln{a}]$

$\displaystyle \frac{y'}{y} = \ln{a}$

$\displaystyle y' = \ln{a} \cdot y = \ln{a} \cdot a^x$

6. With regard to your second question: $\displaystyle \frac{d}{dx}\frac{4^xx^{1/3}}{tan(x)}$, you want to use the product and quotient rules.

$\displaystyle \frac{d}{dx} = \frac{tan(x)*\frac{d}{dx}[4^xx^{1/3}]-4^xx^{1/3}*\frac{d}{dx}[tan(x)]}{tan^2(x)}$

$\displaystyle \frac{d}{dx}[4^xx^{1/3}] = 4^x*\frac{1}{3}x^{-\frac{2}{3}}+x^{\frac{1}{3}}*ln(4)*4^x$

$\displaystyle \frac{d}{dx}[tan(x)] = sec^2(x)$

Plugging this all yields: $\displaystyle \frac{d}{dx} = \frac{tan(x)*(4^x*\frac{1}{3}x^{-\frac{2}{3}}+x^{\frac{1}{3}}*ln(4)*4^x)-4^xx^{1/3}*sec^2(x)}{tan^2(x)}$

I'll leave the simplification to you.

7. wow for some reason I had to keep staring at it but it sunk in.

Thanks to all!