# Math Help - Implicit Differentiation

1. ## Implicit Differentiation

Find dy/dx for:

ln xy = 7 + x + y

I don't know at all where to start.

Find the slope of:

y = (5e^x)^2 when x = Square root of e

2. Originally Posted by Dickson
Find dy/dx for:

ln xy = 7 + x + y

I don't know at all where to start.
Recall that $\ln ab=\ln a+\ln b$

So you need to differentiate $\ln x+\ln y=7+x+y$

Differentiating both sides with respect to x, we have

$\frac{1}{x}+\frac{1}{y}\frac{\,dy}{\,dx}=1+\frac{\ ,dy}{\,dx}$

Can you continue and find $\frac{\,dy}{\,dx}$?

Find the slope of:

y = (5e^x)^2 when x = Square root of e
This is the same as $y=25e^{2x}$. Differntiating, you have $\frac{\,dy}{\,dx}=50e^{2x}$. Now what is $y^{\prime}\!\left(\sqrt{e}\right)$?

Does this make sense?

3. SOrry for the first one I'm still a little lost on how to get the answer

And the second one how did we end up with 50? Is it bringing the exponent down?

4. Originally Posted by Dickson
SOrry for the first one I'm still a little lost on how to get the answer
That's ok. This implicit differentiation stuff can be kind of confusing at first. Note that $y$ is a function of x, as in $y=y\!\left(x\right)$. Now when we differentiate $y\!\left(x\right)$, we do so by differentiating with respect to x. So we note that $\frac{\,d}{\,dx}\left[y\!\left(x\right)\right]$ is analagous to $\frac{\,dy}{\,dx}$.

Now, when we differentiate functions that contain $y$ terms, we need to apply the chain rule, since after all, $y$ is a function of x. For example, when we differentiate $\ln y$, we know that by chain rule, this is the same as $\frac{1}{y}\cdot\frac{\,d}{\,dx}\left[y\right]$. Now above, we mentioned that $\frac{\,d}{\,dx}\left[y\right]$ is the same as $\frac{\,dy}{\,dx}$. So it follows then that $\frac{\,d}{\,dx}\left[\ln y\right]=\frac{1}{y}\cdot\frac{\,d}{\,dx}\left[y\right]=\frac{1}{y}\frac{\,dy}{\,dx}$.

Try to apply this idea to other functions of $y$.

Does this clarify things?

And the second one how did we end up with 50? Is it bringing the exponent down?
In an sense, yes. What you really end up multiplying by is the derivative of the exponent. Its more of a chain rule application: $\frac{\,d}{\,dx}\left[e^{ax}\right]=e^{ax}\cdot\frac{\,d}{\,dx}\left[ax\right]=ae^{ax}$.

So when I differentiated $y=25e^{2x}$, it turned out to be $\frac{\,dy}{\,dx}=25e^{2x}\cdot\frac{\,d}{\,dx}\le ft[2x\right]=50e^{2x}$

Does this demystify things?

5. Yes thank you very much