Question

Find `dy/dx` of `y=(5-x)^3(4+x)^5`?

I understand that I need to use the chain rule, but the x in the parentheses has no power and it is throwing me off

Answer 1

`dy/dx=5(5-x)^3(4+x)^4-3(4+x)^5(5-x)^2`

Explanation:

`y=(5-x)^3(4+x)^5`

`dy/dx=d/dx[(5-x)^3(4+x)^5]`

`color(white)(dy/dx)=(5-x)^3d/dx[(4+x)^5]+(4+x)^5d/dx[(5-x)^3]`

`color(white)(dy/dx)=(5-x)^3(5* (4+x)^(5-1 ) * d/dx[4+x])+(4+x)^5(3*(5-x)^(3-1)*d/dx[5-x])`

`color(white)(dy/dx)=(5-x)^3(5(4+x)^4(1))+(4+x)^5(3(5-x)^2(-1))`

`color(white)(dy/dx)=5(5-x)^3(4+x)^4-3(4+x)^5(5-x)^2`

Answer 2

`dy/dx= 5(5 - x)^3(4 + x)^4 - 3(5 - x)^2(4 + x)^5`

Explanation:

Here is a different way that I personally like to use on these sorts of questions.

Taking the natural logarithm of both sides, we get:

`lny = ln(5 - x)^3(4 + x)^5`

Now recall your logarithm laws. The most important ones here are `ln(ab) = ln(a) + ln(b)` and `ln(a^n) = nlna`

`lny = ln(5 - x)^3 + ln(4 + x)^5`

`lny = 3ln(5 -x) + 5ln(4 + x)`

Now differentiate using the chain rule and the fact that `d/dx(lnx) = 1/x`. Don't forget that you need to differentiate the left hand side with respect to `x`.

`1/y(dy/dx) = -3/(5 - x) + 5/(4 + x)`

`dy/dx= y(5/(4 + x) - 3/(5 - x))`

`dy/dx= (5 - x)^3(4 + x)^5(5/(4 + x) - 3/(5 - x))`

`dy/dx= 5(5 - x)^3(4 + x)^4 - 3(5 - x)^2(4 + x)^5`

Which is the result obtained by the other contributor using the chain rule exclusively.

Hopefully this helps!