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₯) sec π₯.

02:09

### Video Transcript

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

The question wants us to differentiate π of π₯ which is the product of two functions. To differentiate the product of two functions, we recall the product rule for differentiation, which tells us if π’ and π£ are both functions of π₯, then the derivative of π’ times π£ with respect to π₯ is equal to π’ prime π£ plus π’π£ prime. So to differentiate our function π of π₯, weβll set π’ equal to π to the power of π₯ and π£ equal to the sec of π₯. Then, to find our function π’ prime, we need to differentiate π to the power of π₯. And we recall the derivative of π to the power of π₯ with respect to π₯ is just equal to itself. So π’ prime is equal to π to the power of π₯.

Next, to find π£ prime, we need to differentiate the sec of π₯. We recall a standard rule for trigonometric derivatives. The derivative of the sec of π₯ with respect to π₯ is equal to the sec of π₯ times the tan of π₯. You could prove this by writing the sec of π₯ as one divided by the cos of π₯ and then using the quotient rule for differentiation. However, itβs useful to commit this to memory. Using this, we see that π£ prime is equal to the sec of π₯ times the tan of π₯.

Weβre now ready to apply the product rule to find the derivative of π of π₯. Itβs equal to π’ prime π£ plus π’π£ prime. Substituting the functions we found for π’, π£, π’ prime, and π£ prime, we have that π prime of π₯ is equal to π to the power of π₯ times the sec of π₯ plus π to the power of π₯ times the sec of π₯ times the tan of π₯. And we could leave our answer like this. However, we can simplify our answer by noticing that both terms share a factor of π to the π₯ and both terms share a factor of the sec of π₯.

So we take out our factor of π to the power of π₯ times the sec of π₯. We then see that we need to multiply this by one plus the tan of π₯. And this gives us our final answer. If π of π₯ is equal to π to the power of π₯ times the sec of π₯, then the derivative of π of π₯, π prime of π₯, is equal to π to the power of π₯ times the sec of π₯ multiplied by one plus the tan of π₯.

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