# Video: APCALC05AB-P1A-Q16-578160263482

Find d𝑦/d𝑥 if 𝑦 = log₄ (3𝑥² +2𝑥⁴).

03:17

### Video Transcript

Find d𝑦 by d𝑥 if 𝑦 is equal to the log to the base four of three 𝑥 squared plus two 𝑥 to the power of four.

We’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 𝑥 is one over 𝑥, where the natural log ln or ln 𝑥 is log to the base 𝑒 of 𝑥. And since we know the derivative of the natural log of 𝑥, to find the derivative of a log which is not to the base 𝑒, 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 𝑎 of 𝑢 which we’d like to convert to a logarithm of base 𝑏, then log to the base 𝑎 of 𝑢 is equal to log to the base 𝑏 of 𝑢 over log to the base 𝑏 of 𝑎.

In our case, 𝑢 is equal to three 𝑥 squared plus two 𝑥 to the power of four. 𝑎 is equal to four. And 𝑏, our new base, is equal to Euler’s number 𝑒. That’s the base of a natural logarithm. So that by the base conversion identity, our log to the base four of three 𝑥 squared plus two 𝑥 to the power of four is equal to log to the base 𝑒 of three 𝑥 squared plus two 𝑥 to the four over log to the base 𝑒 of four. And using the notation that log to the base 𝑒 of 𝑥 is ln or ln of 𝑥, we have 𝑦 equal to ln or ln of three 𝑥 squared plus two 𝑥 to the power of four over ln or ln four.

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

With our substitution 𝑢 is three 𝑥 squared plus two 𝑥 to the four, we have 𝑦 equal to one over the natural log of four times the natural log of 𝑢. And as we saw earlier, the derivative of the natural log of 𝑥 is one over 𝑥 so that d𝑦 by d𝑢 is one over the natural log of four times one over 𝑢. And going back to our substitution 𝑢 is equal to three 𝑥 squared plus two 𝑥 to the four, that’s one over the natural log of four times one over three 𝑥 squared plus two 𝑥 to the four.

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