Video: APCALC05AB-P1A-Q16-578160263482

Find d𝑦/dπ‘₯ if 𝑦 = logβ‚„ (3π‘₯Β² +2π‘₯⁴).

03:17

Video Transcript

Find d𝑦 by dπ‘₯ if 𝑦 is equal to the log to the base four of three π‘₯ squared plus two π‘₯ 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 π‘₯ is one over π‘₯, where the natural log ln or ln π‘₯ is log to the base 𝑒 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 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 π‘Ž of 𝑒 which we’d like to convert to a logarithm of base 𝑏, then log to the base π‘Ž of 𝑒 is equal to log to the base 𝑏 of 𝑒 over log to the base 𝑏 of π‘Ž.

In our case, 𝑒 is equal to three π‘₯ squared plus two π‘₯ to the power of four. π‘Ž 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 π‘₯ squared plus two π‘₯ to the power of four is equal to log to the base 𝑒 of three π‘₯ squared plus two π‘₯ to the four over log to the base 𝑒 of four. And using the notation that log to the base 𝑒 of π‘₯ is ln or ln of π‘₯, we have 𝑦 equal to ln or ln of three π‘₯ squared plus two π‘₯ to the power of four over ln or ln four.

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

With our substitution 𝑒 is three π‘₯ squared plus two π‘₯ to the four, we have 𝑦 equal to one over the natural log of four times the natural log of 𝑒. And as we saw earlier, the derivative of the natural log of π‘₯ is one over π‘₯ so that d𝑦 by d𝑒 is one over the natural log of four times one over 𝑒. And going back to our substitution 𝑒 is equal to three π‘₯ squared plus two π‘₯ to the four, that’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 d𝑦 by dπ‘₯. d𝑦 by dπ‘₯ is d𝑦 by d𝑒 times d𝑒 by dπ‘₯. And that’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 d𝑦 by dπ‘₯ is equal to eight π‘₯ squared plus six over two π‘₯ cubed plus three π‘₯ times the natural log of four.

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