Derivative Question

KAVIV

New member
Joined
Dec 17, 2014
Messages
1
When taking the derivative of y = (x2/3 + 2e-9x)6
Its a chain rule derivative so the first part is 6 (x2/3 + 2e-9x)5
The first part of the inner function is a simple power rule (2/3) x-1/3
Im stuck on the second part, the x is an exponent so I think to use the exponent rule.
Applying dy/dx bx = bxln (b) I get 2e9x ln (2e). ln (2e) = 1.693 so I get 3.39e9x
The answer is -18-9x.
Where did I go wrong?

Thanks!​
 
The second part looks like 2 * e-9x so the derivative is 2 times the derivative of e-9x. Or, using the exponent rule
\(\displaystyle \frac{d e^{g(x)}}{dx} = ln(e) e^{g(x)} \frac{d g}{dx} = e^{g(x)} \frac{d g}{dx}\)
we have
\(\displaystyle \frac{d e^{-9x}}{dx} = -9 e^{-9x}\)
 
I agree with Ishuda. The function is NOT "\(\displaystyle (2e)^{-9x}\)" as you seem to think, but "\(\displaystyle 2e^{-9x}\)" which has derivative \(\displaystyle 2(-9e^{-9x})= -18e^{-9x}\).
 
Last edited:
It's just more of the chain rule

I agree with Ishuda too. If you already know that d/dx(e^x)=e^x, then the derivative of e^-9x is just an application of the chain rule. You can ignore the 2 in front when taking the derivative for any function.

To really understand why the chain rule works, check out this video: http://youtu.be/pzYHltWq_z8
 
then the derivative of > > > e^-9x < < <

Be careful. You need grouping symbols for \(\displaystyle \ e^{-9x} \ \) when typed horizontally, such as: "e^(-9x)."


Else, what you have is \(\displaystyle \ e^{-9}x.\)
 
Top