Question Video: Differentiating Logarithmic Functions Using the Chain Rule | Nagwa Question Video: Differentiating Logarithmic Functions Using the Chain Rule | Nagwa

Question Video: Differentiating Logarithmic Functions Using the Chain Rule Mathematics

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.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

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