### Video Transcript

Find dπ¦ by dπ₯ if π¦ is equal to
the log to the base four of three π₯ squared plus two π₯ to the power of four.

Weβre asked to find the derivative
of a log function where the log is to the base four. Now we know that the derivative of
the natural log of π₯ is one over π₯, where the natural log ln or ln π₯ is log to
the base π of π₯. And since we know the derivative of
the natural log of π₯, to find the derivative of a log which is not to the base π,
we first convert our log to a natural logarithm. To do this, we use the base
conversion identity. If we have a log to the base π of
π’ which weβd like to convert to a logarithm of base π, then log to the base π of
π’ is equal to log to the base π of π’ over log to the base π of π.

In our case, π’ is equal to three
π₯ squared plus two π₯ to the power of four. π is equal to four. And π, our new base, is equal to
Eulerβs number π. Thatβs the base of a natural
logarithm. So that by the base conversion
identity, our log to the base four of three π₯ squared plus two π₯ to the power of
four is equal to log to the base π of three π₯ squared plus two π₯ to the four over
log to the base π of four. And using the notation that log to
the base π of π₯ is ln or ln of π₯, we have π¦ equal to ln or ln of three π₯
squared plus two π₯ to the power of four over ln or ln four.

Weβre asked to find dπ¦ by dπ₯. To do this, we first note that ln
four is actually a constant. So the function weβre actually
differentiating is log of three π₯ squared plus two π₯ to the four. And this is a function of a
function. So weβre going to use the chain
rule. This says that if π¦ is a function
of π’, where π’ is a function of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times
dπ’ by dπ₯. In our case, we let π’ equal to
three π₯ squared plus two π₯ to the four. And to find dπ’ by dπ₯, we use the
power rule. This says that, for π not equal to
negative one, the derivative of π times π₯ raised to the power π is π times π
times π₯ raised to the power π minus one. That is, we multiply by the
exponent and subtract one from the exponent so that dπ’ by dπ₯ is six π₯ plus eight
π₯ to the power of three.

With our substitution π’ is three
π₯ squared plus two π₯ to the four, we have π¦ equal to one over the natural log of
four times the natural log of π’. And as we saw earlier, the
derivative of the natural log of π₯ is one over π₯ so that dπ¦ by dπ’ is one over
the natural log of four times one over π’. And going back to our substitution
π’ is equal to three π₯ squared plus two π₯ to the four, thatβs one over the natural
log of four times one over three π₯ squared plus two π₯ to the four.

Now we have everything we need to
use the chain rule for dπ¦ by dπ₯. dπ¦ by dπ₯ is dπ¦ by dπ’ times dπ’ by dπ₯. And thatβs equal to one over the
natural log of four times one over three π₯ squared plus two π₯ to the four times
six π₯ plus eight π₯ cubed. We have a common factor of π₯ in
both the numerator and the denominator, which gives us eight π₯ squared plus six
over the natural log of four times two π₯ cubed plus three π₯. So dπ¦ by dπ₯ is equal to eight π₯
squared plus six over two π₯ cubed plus three π₯ times the natural log of four.