Question Video: Finding the Derivative of a Function Involving Trigonometric and Exponential Functions Using the Product Rule

Differentiate 𝑓(π‘₯) = 𝑒^(π‘₯) sin π‘₯.

02:04

Video Transcript

Differentiate 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯ times the sin of π‘₯.

The question wants us to differentiate our function 𝑓 of π‘₯. And we can see that our function is the product of two functions. It’s the product of 𝑒 to the π‘₯ and the sin of π‘₯. And we know how to differentiate the product of two functions by using the product rule. The product rule tells us if we have a function 𝑓 of π‘₯ which is the product of two functions 𝑒 of π‘₯ and 𝑣 of π‘₯, then 𝑓 prime of π‘₯ is equal to 𝑣 of π‘₯ times 𝑒 prime of π‘₯ plus 𝑒 of π‘₯ times 𝑣 prime of π‘₯.

And 𝑓 of π‘₯ is the product of two functions. So, we’ll set 𝑒 of π‘₯ to be 𝑒 to the power of π‘₯ and 𝑣 of π‘₯ to be the sin of π‘₯. To use the product rule, we need to find expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯. Let’s start with 𝑒 prime of π‘₯. We see that 𝑒 prime of π‘₯ will be the derivative of 𝑒 to the power of π‘₯ with respect to π‘₯. And we know that this differentiates to give itself. So, 𝑒 prime of π‘₯ is also equal to 𝑒 to the power of π‘₯.

Next, let’s find an expression for 𝑣 prime of π‘₯. That’s the derivative of the sin of π‘₯ with respect to π‘₯. And this is a standard trigonometric derivative result that we should commit to memory. The derivative of the sin of π‘₯ with respect to π‘₯ is equal to the cos of π‘₯. So, 𝑣 prime of π‘₯ is equal to the cos of π‘₯.

We’re now ready to find an expression for our derivative function 𝑓 prime of π‘₯. By the product rule, it’s equal to 𝑣 of π‘₯ times 𝑒 prime of π‘₯ plus 𝑒 of π‘₯ times 𝑣 prime of π‘₯. Substituting our expressions for 𝑒 of π‘₯, 𝑣 of π‘₯, 𝑒 prime of π‘₯, and 𝑣 prime of π‘₯, we get that 𝑓 prime of π‘₯ is equal to the sin of π‘₯ times 𝑒 to the power of π‘₯ plus 𝑒 to the power of π‘₯ times the cos of π‘₯.

We could leave our answer like this; however, we’ll take out the shared factor of 𝑒 to the power of π‘₯. And this gives us 𝑒 to the power of π‘₯ times the sin of π‘₯ plus the cos of π‘₯. And this is our final answer. Therefore, we’ve shown if 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯ times the sin of π‘₯, then 𝑓 prime of π‘₯ is equal to 𝑒 to the power of π‘₯ times the sin of π‘₯ plus the cos of π‘₯.

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