Question

Determine the slope of the graph of x2 = ln(xy) at the point (1, e)
Do I start with implicit differentiation? Or is there another way? I do not know what to do.

Answer 1

@hero please help :)

Answer 2

Try isolating y first.

Answer 3

Afterwards, find y'

Answer 4

so many people

Answer 5

\[y=\frac{ 1 }{ x ^{3} }\]
@Hero So far so good?

Answer 6

You can isolate the y, and as I did that I obtained
\(\large\color{#000000 }{ \displaystyle y(x)=e^{x^2}/x }\)

(Note, if you want to integrate that, then
top and bottom •x, and then set u=x\(^2\),)

Answer 7

9(No, you don't need to integrate this, this is not about integration at all, but just in case you wondered about it))

Answer 8

If you want, you could do implicit differentiation as well.

\(\large\color{#000000 }{ \displaystyle x^2 = \ln(xy) }\)
\(\large\color{#000000 }{ \displaystyle x^2 = \ln(x)+\ln(y) }\)

then, differentiate each component.
(where the derivative of ln(y) deserves a y')

Answer 9

Thanks!

Answer 10

Just in case, (to nail the concept if it's not nailed already).

\(\large\color{#000000 }{ \displaystyle \frac{ d}{dx}\left[~\ln f(x)~\right]=\frac{1}{f(x)}\times f'(x)=\frac{f'(x)}{f(x)} }\)

(This is when you take an "ln" of a function; via the chain rule)

And that above is exactly the equivalent of
\(\large\color{#000000 }{ \displaystyle \frac{ d}{dx}\left[~\ln y~\right]=\frac{1}{y}\times y'=\frac{y'}{y} }\)

Because y is a function of x (just as f(x) is), and therefore, just as f(x) it desrves a chain rule.

((In case you might have been wondering where this y' always comes from; hope I addressed this question))