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

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

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Video Transcript

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

We need to differentiate the function 𝑓 of π‘₯. We can see that 𝑓 is a function of π‘₯. So, we need to differentiate this with respect to π‘₯. To do this, we need to notice that 𝑓 of π‘₯ is the product of two functions. It’s 𝑒 to the power of π‘₯ multiplied by the csc of π‘₯. And in fact, it’s the product of two functions which we know how to differentiate. This means we can do this by using the products rule. So, let’s recall what the product rule tells us.

The product rule tells us if 𝑓 of π‘₯ is equal to the product of two differentiable functions 𝑒 of π‘₯ times 𝑣 of π‘₯, then 𝑓 prime of π‘₯ is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ plus 𝑣 prime of π‘₯ times 𝑒 of π‘₯. To use the product rule in this case, we’re going to need to set 𝑒 of π‘₯ equal to 𝑒 to the power of π‘₯ and 𝑣 of π‘₯ equal to the csc of π‘₯. This then means that 𝑓 of π‘₯ is equal to 𝑒 of π‘₯ multiplied by 𝑣 of π‘₯.

This means to use the product rule, we’re now going to need to find expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯. Let’s start with 𝑒 prime of π‘₯. That’s the derivative of 𝑒 to the power of π‘₯ with respect to π‘₯. And this is a standard derivative result we should commit to memory. The derivative of the exponential function 𝑒 to the power of π‘₯ with respect to π‘₯ is just equal to itself, 𝑒 to the power of π‘₯. So, 𝑒 prime of π‘₯ is just 𝑒 to the power of π‘₯.

We now need to find an expression for 𝑣 prime of π‘₯. That’s the derivative of the csc of π‘₯ with respect to π‘₯. There’s a few different ways we could evaluate this derivative. For example, we could write the csc of π‘₯ as one divided by the sin of π‘₯ and then use either the quotient rule or the general power rule. However, it’s easier to just recall that the derivative of the csc of π‘₯ with respect to π‘₯ is equal to negative the csc of π‘₯ times the cot of π‘₯. So, using this, we now have 𝑣 prime of π‘₯ is equal to negative the csc of π‘₯ times the cot of π‘₯.

Now that we found expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯, we’re ready to find an expression for 𝑓 prime of π‘₯ by using the product rule. It’s equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ plus 𝑣 prime of π‘₯ times 𝑒 of π‘₯. Substituting in 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 𝑒 to the power of π‘₯ times the csc of π‘₯ plus negative one times the csc of π‘₯ multiplied by the cot of π‘₯ times 𝑒 to the power of π‘₯.

And there’s a few ways we can simplify this. We’ll take out the common factor of 𝑒 to the power of π‘₯ and the common factor of the csc of π‘₯. And this leaves us with our final answer, 𝑒 to the power of π‘₯ times the csc of π‘₯ multiplied by one minus the cot of π‘₯.

Therefore, given 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯ times the csc of π‘₯, we were able to use the product rule to show that 𝑓 prime of π‘₯ is equal to 𝑒 to the power of π‘₯ times the csc of π‘₯ multiplied by one minus the cot of π‘₯.

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