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
Recall that $\displaystyle \ln ab=\ln a+\ln b$
So you need to differentiate $\displaystyle \ln x+\ln y=7+x+y$
Differentiating both sides with respect to x, we have
$\displaystyle \frac{1}{x}+\frac{1}{y}\frac{\,dy}{\,dx}=1+\frac{\ ,dy}{\,dx}$
Can you continue and find $\displaystyle \frac{\,dy}{\,dx}$?
This is the same as $\displaystyle y=25e^{2x}$. Differntiating, you have $\displaystyle \frac{\,dy}{\,dx}=50e^{2x}$. Now what is $\displaystyle y^{\prime}\!\left(\sqrt{e}\right)$?Find the slope of:
y = (5e^x)^2 when x = Square root of e
Does this make sense?
That's ok. This implicit differentiation stuff can be kind of confusing at first. Note that $\displaystyle y$ is a function of x, as in $\displaystyle y=y\!\left(x\right)$. Now when we differentiate $\displaystyle y\!\left(x\right)$, we do so by differentiating with respect to x. So we note that $\displaystyle \frac{\,d}{\,dx}\left[y\!\left(x\right)\right]$ is analagous to $\displaystyle \frac{\,dy}{\,dx}$.
Now, when we differentiate functions that contain $\displaystyle y$ terms, we need to apply the chain rule, since after all, $\displaystyle y$ is a function of x. For example, when we differentiate $\displaystyle \ln y$, we know that by chain rule, this is the same as $\displaystyle \frac{1}{y}\cdot\frac{\,d}{\,dx}\left[y\right]$. Now above, we mentioned that $\displaystyle \frac{\,d}{\,dx}\left[y\right]$ is the same as $\displaystyle \frac{\,dy}{\,dx}$. So it follows then that $\displaystyle \frac{\,d}{\,dx}\left[\ln y\right]=\frac{1}{y}\cdot\frac{\,d}{\,dx}\left[y\right]=\frac{1}{y}\frac{\,dy}{\,dx}$.
Try to apply this idea to other functions of $\displaystyle y$.
Does this clarify things?
In an sense, yes. What you really end up multiplying by is the derivative of the exponent. Its more of a chain rule application: $\displaystyle \frac{\,d}{\,dx}\left[e^{ax}\right]=e^{ax}\cdot\frac{\,d}{\,dx}\left[ax\right]=ae^{ax}$.And the second one how did we end up with 50? Is it bringing the exponent down?
So when I differentiated $\displaystyle y=25e^{2x}$, it turned out to be $\displaystyle \frac{\,dy}{\,dx}=25e^{2x}\cdot\frac{\,d}{\,dx}\le ft[2x\right]=50e^{2x}$
Does this demystify things?