Asked by Nicky

Derivative of:

(exp(8x^4))(5x^3+9)^3
___________________________ (divided by)

(22x^2+4x-20)^2

Answers

Answered by Helper
What is exp?
Answered by Nicky
exp is euler's constant around 2.718.......
Answered by Helper
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.
Answered by Nicky
e^(8x^4)
Answered by Reiny
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.

Answered by Nicky
yes, i understand that part but i do not know how to actually solve the problem
Answered by Reiny
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.
Answered by Nicky
i am confused by the bracket.. there is only 1 bracket
Answered by Helper
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.
Answered by Helper
Like I said, more than one way to do this.

I suggest follow Reiny's way.
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