### Video Transcript

Find dπ¦ by dπ₯ if five π¦π to the
two π₯ equals seven π to the five.

Now, at first glance, this does
look a little complicated. However, we can clearly see that we
can rearrange the equation to make π¦ the subject. Weβre going to divide both sides of
the equation by five π to the two π₯. On the left-hand side, that leaves
us simply with π¦. And on the right-hand side, we have
seven π to the power of five over five π to the two π₯. Now, actually, one over π to the
power of two π₯ is equal to π to the negative two π₯. So we can rewrite our equation. And we say that π¦ is equal to
seven π to the power of five over five times π to the negative two π₯.

Notice that seven π to the five
over five is just a constant. So we can differentiate this using
the general formula for the derivative of the exponential function. The derivative of π to the ππ₯
with respect to π₯ is ππ to the ππ₯. And of course, remembering that the
constant factor rule allows us to take constants outside a derivative and
concentrate on differentiating the function of π₯ itself.

In other words, dπ¦ by dπ₯ is equal
to seven π to the power of five over five times the derivative of π to the
negative two π₯ with respect to π₯. And the derivative of π to the
negative two π₯ with respect to π₯ is negative two π to the negative two π₯. This means dπ¦ by dπ₯ is negative two
times seven π to the power of five over five times π to the negative two π₯.

Notice that the derivative can
actually be expressed in terms of π¦ since we said that π¦ was equal to seven π to
the power of five over five times π to the negative two π₯. And we can therefore say that dπ¦
by dπ₯ is equal to negative two π¦.