# Video: APCALC05AB-P1A-Q16-578160263482

Find dπ¦/dπ₯ if π¦ = logβ (3π₯Β² +2π₯β΄).

03:17

### 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.

Weβre 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 weβd 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 Eulerβs number π. Thatβs 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.

Weβre asked to find dπ¦ by dπ₯. To do this, we first note that ln four is actually a constant. So the function weβre actually differentiating is log of three π₯ squared plus two π₯ to the four. And this is a function of a function. So weβre 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, thatβs 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 thatβs 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.