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

# Question Video: Finding the Derivative of a Function Involving Trigonometric and Exponential Functions Using the Product Rule Mathematics • Third Year of Secondary School

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Differentiate π(π₯) = π^(π₯) sin π₯.

02:04

### Video Transcript

Differentiate π of π₯ is equal to π to the power of π₯ times the sin of π₯.

The question wants us to differentiate our function π of π₯. And we can see that our function is the product of two functions. Itβs the product of π 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 π of π₯ which is the product of two functions π’ of π₯ and π£ of π₯, then π prime of π₯ is equal to π£ of π₯ times π’ prime of π₯ plus π’ of π₯ times π£ prime of π₯.

And π of π₯ is the product of two functions. So, weβll set π’ of π₯ to be π to the power of π₯ and π£ of π₯ to be the sin of π₯. To use the product rule, we need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. We see that π’ prime of π₯ will be the derivative of π to the power of π₯ with respect to π₯. And we know that this differentiates to give itself. So, π’ prime of π₯ is also equal to π to the power of π₯.

Next, letβs find an expression for π£ prime of π₯. Thatβ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, π£ prime of π₯ is equal to the cos of π₯.

Weβre now ready to find an expression for our derivative function π prime of π₯. By the product rule, itβs equal to π£ of π₯ times π’ prime of π₯ plus π’ of π₯ times π£ prime of π₯. Substituting our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯, we get that π prime of π₯ is equal to the sin of π₯ times π to the power of π₯ plus π to the power of π₯ times the cos of π₯.

We could leave our answer like this; however, weβll take out the shared factor of π to the power of π₯. And this gives us π to the power of π₯ times the sin of π₯ plus the cos of π₯. And this is our final answer. Therefore, weβve shown if π of π₯ is equal to π to the power of π₯ times the sin of π₯, then π prime of π₯ is equal to π to the power of π₯ times the sin of π₯ plus the cos of π₯.

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