Derivative of f(x) = ln(2 - x - 5x^2)

[seaborg]

when trying to solve this problem I applied the chain rule to obtain

dy= dy X du ; u = (2 - x - 5x^2)
dx ..du ...dx

-1 - 10x_ X ( -1 - 10x )
(2 - x - 5x^2)

but WolframAlpha and my textbook agree that the solution is complete without multiplying by the derivative of the contents of ln()

-1 - 10x___
(2 - x - 5x^2)

 

[daon2]

dy/du is 1/u, not u'/(u). You are mistakenly multiplying by u' twice.

 

[srmichael]

when trying to solve this problem I applied the chain rule to obtain

dy= dy X du ; u = (2 - x - 5x^2)
dx ..du ...dx

-1 - 10x_ X ( -1 - 10x )
(2 - x - 5x^2)

but WolframAlpha and my textbook agree that the solution is complete without multiplying by the derivative of the contents of ln()

-1 - 10x___
(2 - x - 5x^2)

The derivative of ln(u) is u'/u.

So in your case it is just (-1-10x)/(2-x-5x²)

 

[HallsofIvy]

What they said! In a little more detail, you are letting \(\displaystyle u= 2- x- 5x^2\) so that y(u)= ln(u). Then \(\displaystyle \frac{dy}{du}= \frac{1}{u}\) and \(\displaystyle \frac{du}{dx}= -1- 10x\). So, by the chain rule \(\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= \frac{1}{u}(-1- 10x)= \frac{-1- 10x}{2- x- 5x^2}\). Your error was not realizing that you had already included the "\(\displaystyle \frac{du}{dx}\)" in what you were thinking was just \(\displaystyle \frac{dy}{du}\).

 

[Deleted member 4993]

when trying to solve this problem I applied the chain rule to obtain

dy= dy X du ; u = (2 - x - 5x^2)
dx ..du ...dx

-1 - 10x_ X ( -1 - 10x )
(2 - x - 5x^2)

but WolframAlpha and my textbook agree that the solution is complete without multiplying by the derivative of the contents of ln()

-1 - 10x___
(2 - x - 5x^2)

Do not use X - to indicate multiplication. Use '_*' instead