Video Transcript
Determine the derivative of π of
π₯ is equal to two times the square root of two π₯ minus one.
The question wants us to find an
expression for the derivative of our function π of π₯. And we can see that π of π₯ is two
times the square root of two π₯ minus one. And the square root of two π₯ minus
one is the composition of two functions. Weβre taking the square root of a
linear function. So weβre going to want to use the
chain rule to evaluate this derivative. We recall the following version of
the chain rule. If we have a function π of π₯
which is the composition of two functions π’ and π£, then the chain rule tells us π
prime of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯. So letβs write our function π of
π₯ in this form.
Our innermost function is the
linear function two π₯ minus one. So weβll set π£ of π₯ to be equal
to two π₯ minus one. Using this, we have π of π₯ is
equal to two times the square root of π£ of π₯. We see weβre taking the square root
of π£ of π₯ and then multiplying this by two. So to write π of π₯ as the
composition of π’ and π£, weβll need π’ evaluated at π£ to be equal to two times the
square root of π£. In other words, for these
functions, π’ and π£, we have that π of π₯ is equal to π’ evaluated at π£ evaluated
at π₯, which is two times the square root of two π₯ minus one. So we can now apply the chain rule
to find π prime of π₯.
To use the chain rule, we need to
find π£ prime and π’ prime. Letβs start by finding π£ prime of
π₯. Thatβs the derivative of two π₯
minus one with respect to π₯. And of course, this is a linear
function, so its slope is equal to the coefficient of π₯, which is two. We couldβve also evaluated this by
using the power rule for differentiation. Letβs now find an expression for π’
prime of π£. Thatβs the derivative of two times
the square root of π£ with respect to π£. We can do this by using our laws of
exponents. We know that two root π£ is equal
to two times π£ to the power of one-half.
We can then differentiate this by
using the power rule for differentiation. We multiply by the exponent of π£,
thatβs one-half, and then reduce this exponent by one. This gives us one-half times two
times π£ to the power of one-half minus one. We can simplify this
expression. One-half times two is equal to one,
and one-half minus one is equal to negative one-half. So this gives us that π’ prime of
π£ is equal to π£ to the power of negative one-half. Using our laws of exponents, we can
write this as one divided by the square root of π£. So weβve now found expressions for
π£ prime of π₯ and π’ prime of π£. This means weβre now ready to use
our chain rule to evaluate π prime of π₯.
The chain rule tells us π prime of
π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯. We showed that π£ prime of π₯ is
equal to two. And we showed that π’ prime of π£
is equal to one divided by the square root of π£. Finally, we want to use our
substitution π£ is equal to two π₯ minus one. This gives us that π prime of π₯
is equal to two times one divided by the square root of two π₯ minus one, which we
can simplify to be two divided by the square root of two π₯ minus one. Therefore, weβve shown if π of π₯
is equal to two times the square root of two π₯ minus one, then π prime of π₯ is
equal to two divided by the square root of two π₯ minus one.
