💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Use logarithmic differentiation to find the derivative of the function.$y = x^{\sin x}$

## $$y^{\prime}=x^{\sin x}\left(\frac{\sin x}{x}+\ln x \cos x\right)$$

Derivatives

Differentiation

### Discussion

You must be signed in to discuss.
##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp

### Video Transcript

So for this problem we have is right equals X to the sign of acts. And whenever we're dealing with these kinds of functions where we have an exponents with the X and the base with the ex, we want to implement logarithms Andi specifically the natural log. Here's why If we take the natural log of both sides, we can simplify this by getting rid of the exponents. So now we have a sine X out in front. Then we can take the derivative of this and perform implicit differentiation. So when we take the derivative of this portion right here, we get one over. Why times why prime? And that's going to be equal Thio here. What? You will use what's known as the product rule. You may have already been doing it a lot a tous point. So if the product role we take the first portion of the product and just keep it there but we multiply it by the derivative of the second portion. So this portion right here, natural log of X, give us one over X and then what we have is plus, we keep the natural log of access time and then multiply it by the derivative of Sine X, which is just Code Synnex. Now that we've done this, we want Thio, Combine all these terms and simplify further. So we'll get the sign of X over X and then this will just become natural log of X code Synnex Um And then lastly, what we want to do is multiply the Y on both sides. So when we do that, we have Why over here? But we know that why is ultimately equal to X to the cynics. So that's gonna be our final answer. Notice how we make things a little bit more messy by putting the natural logs in. But ultimately it made things simpler because it allowed us to do product rule rather than having to deal with the exponents, which are much more difficult to differentiate. Um, so it's a helpful technique that you will use oftentimes, whenever you see an exponents with a variable in both the base and the exponents portion

California Baptist University

#### Topics

Derivatives

Differentiation

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp