Derivative of:
(exp(8x^4))(5x^3+9)^3
___________________________ (divided by)
(22x^2+4x-20)^2
What is exp?
exp is euler's constant around 2.718.......
Just making sure. Do you mean
e^(8x^4) or e(8x^4)
I've never see "e" referred to "exp" on this site, but I haven't been here long.
e^(8x^4)
in some computer languages the function
y = e^x would be written as
y = exp(x)
I know BASIC and I believe VISUAL BASIC still function that way.
yes, i understand that part but i do not know how to actually solve the problem
My previous reply was only meant for "helper"
As to your question ......
Here is a little trick you might use
let y = e^(8x^4)(5x^3 + 9)^3 / (22x^2 + 4x - 20)^2
take ln of both sides
lny = ln(e^(8x^4) + ln(5x^3 + 9)^3 - ln(22x^2 + 4x - 20)^2
= 8x^4 + 3ln(5x^3 + 9) - 2ln (22x^2 + 4x - 20)
now differentiate with respect to x
(dy/dx)/y = 32x^3 + 3(15x^2)/(5x^3 + 9) - 2(44x + 4)/(22x^2 + 4x - 20)
dy/dx = y [32x^3 + 3(15x^2)/(5x^3 + 9) - 2(44x + 4)/(22x^2 + 4x - 20) ]
=(e^(8x^4)(5x^3 + 9)^3 / (22x^2 + 4x - 20)^2) [ 32x^3 + 3(15x^2)/(5x^3 + 9) - 2(44x + 4)/(22x^2 + 4x - 20)
I know that looks very messy, but it was actually easier than trying to do it with a combination of quotient-product-chain rule.
I also suggest you check my steps and my typing.
i am confused by the bracket.. there is only 1 bracket
Thanks Reiny.
Nicky, depending on your approach, you need to use both the product rule and the chain rule (more than once).
It's easy to get lost in this problem.
What exactly do you not understand?
Have you tried setting the problem up?
One approach, start with the product rule,
d/dx (uv) = v du/dx + u dv/dx
u = e^(8x^4)
v = ((5x^3+9)^3)/(22x^2 + 4x - 20)^2
Then use the chain rule,
u = 8x^4
Then the product rule again,
u = 1/(22x^2 + 4x - 20)^2
v = (5x^3 + 9)^3
Then the chain rule again,
u = (22x^2 + 4x - 20)
and so on.
There might be an another, easier approach. Give it a try and post with specific questions.
Like I said, more than one way to do this.
I suggest follow Reiny's way.