Question

Differentiate the following w.r.t to x. (working included)

Answer 1

The question that needs to be differentiated is this one. I'll write down my working in a second!

\[y^x = x\]

Answer 2

If \[y^x = x / then / y = x^{1/x}\]

\[\ln y = \frac{ 1 }{ x } \ln x\]

\[\frac{ 1 }{ y } \frac{ dx }{dy } = \frac{ 1-\ln x }{ x^2 }\]

to be continued..

Answer 3

\[y*(\frac{ 1-\ln x }{ x^2 }) = \frac{ dx }{dy } = (\frac{ x^{1/x}(1-\ln x )}{ x^2 })\]

Answer 4

However my book says that the answer is:

\[\ln y + \frac{ x }{ y } \frac{ dy }{ dx } = \frac{ 1 }{ x }\]

Where have I gone wrong?

Thanks in advance!

Answer 5

in this step:
\[ \frac{ 1 }{ y } \frac{ dx }{dy } = \frac{ 1-\ln x }{ x^2 } \]
that should be dy/dx on the left side (not dx/dy)

Answer 6

if we let y' stand for dy/dx (for easier typing) , you have
\[ \frac{1}{y} \ y' = \frac{1-\ln x}{x^2} \]
or, (multiply both sides by x, and replace ln x with ln y^x):
\[ \frac{x}{y} \ y' = \frac{1-\ln y^x}{x} \]

Answer 7

and that can be tweaked to match the book's answer

Answer 8

Apologies for the typo!

I see what my book has done, it seems a little bit convoluted, but I suppose I have no choice but to go through with it!

Thanks for the help! c:

Answer 9

yes, there are an infinite number of ways to do the algebra, and I"m not sure why they picked that form.

Answer 10

To me,
x ln y = lnx
then you just take derivative both sides. The LHS is a product rule
(x)' lny + x (lny)'= 1/x
lny + x/y (y)'= 1/x
matches with the book

Answer 11

That makes sense! Perhaps I over complicated the process instead of taking it in the straightforward fashion that you did. Thank you as well!

Answer 12

:)