### Video Transcript

Find the first derivative of
the function π¦ equals the square root of eight π₯ minus sin of nine π₯ to the
power of eight.

Here we have π¦ is equal to the
square root of another function, so we have a composite function. Weβre, therefore, going to
apply the chain rule. Weβre going to define π’ to be
the function underneath the square root, so π’ is equal to eight π₯ minus sin of
nine π₯ to the power of eight. Then, π¦ is equal to the square
root of π’, which we can express using index notation as π’ to the power of
one-half.

The chain rule tells us that
dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯. So, we need to find each of
these derivatives. dπ¦ by dπ’ is relatively straightforward. Using the power rule, we get
one-half π’ to the power of negative one-half. For dπ’ by dπ₯, the derivative
of eight π₯ is just eight. But what about the derivative
of sin of nine π₯ to the power of eight?

We actually need to apply the
chain rule again. We can let π equal this
function. And we can change the notation
a little to write it as sin nine π₯ to the power of eight. Itβs an equivalent notation,
but it might it make it a little clearer how weβre going to find the
derivative.

We recall the chain rule
extension to the power rule, which told us that if we had a function π of π₯
raised to a power π, then its derivative was π prime of π₯ multiplied by π
multiplied by π of π₯ to the power of π minus one. Here, we have a function, sin
of nine π₯ raised to a power eight, so we can apply the chain rule extension to
the power rule. We need to recall one more rule
which is that the derivative with respect to π₯ of sin ππ₯ is π cos ππ₯.

So, we begin. The derivative of the part
inside the parentheses is nine cos nine π₯. Then, we multiply by the power
eight. And then, we have the function
inside the parentheses written out again, but with the power reduced by one. Simplifying gives 72 cos nine
π₯ sin to the power of seven nine π₯. So, now that we found both dπ¦
by dπ’ and dπ’ by dπ₯, we can substitute into the chain rule.

We have then that dπ¦ by dπ₯ is
equal to a half π’ to power of negative a half multiplied by eight minus 72 cos
nine π₯ sin nine π₯ to the power of seven. We must also remember to
replace π’ in terms of π₯. So, π’ is equal to eight π₯
minus sin of nine π₯ to the power of eight. Weβll also simplify the
fractions. Dividing by that denominator of
two leaves coefficients of four and 36 in the numerator.

And we recall also that π’ to
the power of negative a half is equal to one over root π’. So, our derivative dπ¦ by dπ₯
simplifies to four minus 36 cos nine π₯ sin nine π₯ to the power of seven all
over the square root of eight π₯ minus sin nine π₯ to the power of eight.