Question

How do I find the derivative of 3^(4x)?

Answer 1

Hey there :)
Here is our exponential derivative as a reminder:\[\large\rm \frac{d}{dx}a^x=a^x(\ln a)\]

Answer 2

For our problem here,
since the exponent is more than simply x,
we have one extra step, `chain rule`.

Answer 3

If you haven't learned your exponential derivative yet,
then it's possible your teacher wants you to use `logarithmic differentiation`.
Have you seen the previously mentioned rule before? :)

Answer 4

Therefore the derivative would be 4ln(3)x3^(4x)?

Answer 5

Hmm looks very close.
I'm not sure about that x squished in the middle.\[\large\rm \frac{d}{dx}3^{4x}\quad=\quad 3^{4x}(\ln3)(4x)'\]\[\large\rm \frac{d}{dx}3^{4x}\quad=\quad 3^{4x}(\ln3)(4)\]

Answer 6

Suppose you have,
\(\color{#0000ff}{ \displaystyle y=a^x }\)

Take the natural-log of both sides.
\(\color{#0000ff}{ \displaystyle \ln y=\ln a^x }\)

The right side simplifies to.
\(\color{#0000ff}{ \displaystyle \ln y=x(\ln a) }\)

When you differentiate, remember that:
\(\color{}{ [1]\quad d/dx (\ln x)=1/x }\)
\(\color{}{ [2]\quad d/dx (\ln f(x))=f'(x)/f(x) }\) (in [2], Chain Rule is applied)

(You need a chain rule for y, since it's a function of x, just as f(x) is,
and x(ln a), is just x multiplied times some constant ln(a). )

When differentiating both sides, you get:
\(\color{#0000ff}{ \displaystyle \frac{y'}{y}=\ln(a) }\)

Multiply both sides times y.
\(\color{#0000ff}{ \displaystyle y'=y\cdot\ln(a) }\)

You know from the beginning that, \(\color{}{ y=a^x }\). Plug that!
\(\color{#0000ff}{ \displaystyle y'=a^x\cdot\ln(a) }\)

Therefore, you can conclude that:
\(\color{#ff0000}{ \displaystyle \frac{d}{dx}(a^x)=a^x\ln(a) }\)

*********************************************************
Suppose you had,
\(\color{#0000ff}{ \displaystyle y=a^{g(x)} }\)

Then, you can apply the same logarithmic differentiation....
\(\color{#0000ff}{ \displaystyle \ln y=g(x)\cdot \ln (a ) \quad \Longrightarrow \quad y'/y=a\cdot g'(x) \quad \Longrightarrow \quad y'=y\cdot a\cdot g'(x) }\)
\(\color{#0000ff}{ \displaystyle \Longrightarrow \quad y'=a^{g(x)} \cdot a\cdot g'(x) }\)

So, you may conclude:
\(\color{#ff0000}{ \displaystyle \frac{d}{dx}(a^{g(x)})=a^{g(x)}g'(x)\ln(a) }\)

Answer 7

I would never actually do these proves myself, this is just differentiation, except that you don't know what the function and the constants are. (I do proves in Physics, because I am bad at it.)