Video: Differentiating Logarithmic Functions Using the Chain Rule

Find the first derivative of the function 𝑦 = ln (βˆ’5π‘₯⁴ + 2π‘₯Β²).

03:28

Video Transcript

Find the first derivative of the function 𝑦 equals ln of negative five π‘₯ to the power of four plus two π‘₯ squared.

So this is the function we’ve been asked to differentiate, which is in fact a function of a function. And the way to differentiate a function of a function is by using the chain rule. The chain rule says that if 𝑦 equals 𝑓 of 𝑒 and 𝑒 equals 𝑔 of π‘₯ then d𝑦 by dπ‘₯ equals d𝑦 by d𝑒 multiplied by d𝑒 by dπ‘₯. And for our question, 𝑒 is the inner function negative five π‘₯ to the power of four plus two π‘₯ squared. And 𝑦 is the outer function ln 𝑒.

We can see from the formula for the chain rule that we’re going to need d𝑦 by d𝑒 and d𝑒 by dπ‘₯. So, let’s find d𝑦 by d𝑒 first. Well, 𝑦 equals ln of 𝑒. So, we’re gonna differentiate this with respect to 𝑒. And to do this, we recall the general rule that tells us if 𝑦 equals ln of π‘₯, then d𝑦 by dπ‘₯ equals one over π‘₯. And so for 𝑦 equals ln of 𝑒, d𝑦 by d𝑒 equals one over 𝑒.

Okay, so now, we need to find d𝑒 by dπ‘₯. And remember that we have that 𝑒 is equal to negative five π‘₯ to the power of four plus two π‘₯ squared. And before we differentiate this, let’s just remember the power rule of differentiating. That is, if 𝑦 equals π‘Žπ‘₯ to the power of 𝑛, d𝑦 by dπ‘₯ equals π‘›π‘Žπ‘₯ to the power of 𝑛 minus one. And so, with this in mind, d𝑒 by dπ‘₯ is equal to negative 20π‘₯ to the power of three plus four π‘₯.

And so now, applying the formula for the chain rule, d𝑦 by dπ‘₯ equals one over 𝑒 multiplied by negative 20π‘₯ to the power of three plus four π‘₯, which we can write as a single fraction. And we also remember that we already defined 𝑒 as negative five π‘₯ to the power of four plus two π‘₯ squared.

And to tidy this up, since we have a negative here and here, we’ll multiply by negative one over negative one to get 20π‘₯ to the power of three minus four π‘₯ over five π‘₯ to the power of four minus two π‘₯ squared. And we can actually take out π‘₯ as a common factor in both the numerator and the denominator. Which means we can cancel π‘₯ on the top and bottom. And that gives us 20π‘₯ squared minus four over five π‘₯ to the power of three minus two π‘₯. And for our final answer, we’ll take out π‘₯ as a common factor on the denominator. And there’s our final answer.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.