logarithmic derivative of x <mrow class="MJX-TeXAtom-ORD"> e

Martin Nunez

Martin Nunez

Answered question

2022-06-01

logarithmic derivative of x e ( x 2 + cos x )
I'm having a hard time taking the derivative of
f ( x ) = x e ( x 2 + c o s x ) .
I'm aware that I have to take the logarithm of both sides.
ln ( y ) = ln ( x e ( x 2 + cos x ) ) = ln ( x ) e ( x 2 + cos x )
Which I tried to untie, so lets start: First I use the product rule:
1 y y = 1 x e ( x 2 + cos x ) + ln ( x ) e ( x 2 + cos x )
Next the power rule:
1 y y = 1 x e ( x 2 + cos x ) + ln ( x ) e ( x 2 + c o s x ) x 2 + cos x e ( x 2 + cos x ) 1
Then I bring the y to the right:
y = y ( 1 x e ( x 2 + cos x ) + ln ( x ) e ( x 2 + cos x ) x 2 + cos x e ( x 2 + cos x ) 1 2 x sin x )
And exchange y with the term:
y = x e ( x 2 + cos x ) ( 1 x e ( x 2 + cos x ) + ln ( x ) e ( x 2 + cos x ) x 2 + cos x e ( x 2 + cos x ) 1 2 x sin x )
This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn't seem to be correct. Any help would be appreciated.

Answer & Explanation

snowbargerpgl1f

snowbargerpgl1f

Beginner2022-06-02Added 3 answers

You're right until writing
ln f ( x ) = ln x e x 2 + cos x .
The rest, I'm afraid, is completely wrong.
You can differentiate both sides using the product rule:
f ( x ) f ( x ) = ( ln x ) e x 2 + cos x + ln x ( e x 2 + cos x )
Let's split the problem, now: what's ( e x 2 + cos x ) ? Here you can apply the chain rule:
( e x 2 + cos x ) = ( 2 x sin x ) e x 2 + cos x .
Therefore you have
f ( x ) f ( x ) = 1 x e x 2 + cos x + ln x ( 2 x sin x ) e x 2 + cos x
Schenone2pare

Schenone2pare

Beginner2022-06-03Added 1 answers

You just need a way to keep track of what you've done so far, and apply the chain rule in steps. Let's start where you did:
ln ( y ) = ln ( x ) e x 2 + cos x
Then the product rule and chain rule give
y y = [ d d x ln x ] e x 2 + cos x + ln ( x ) [ d d x e x 2 + cos ( x ) ] = 1 x e x 2 + cos x + ln ( x ) e x 2 + cos x d d x ( x 2 + cos x ) = 1 x e x 2 + cos x + ln ( x ) e x 2 + cos x ( 2 x sin x ) .

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