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‘₯) sec π‘₯.

02:09

Video Transcript

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

The question wants us to differentiate 𝑓 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 𝑒 and 𝑣 are both functions of π‘₯, then the derivative of 𝑒 times 𝑣 with respect to π‘₯ is equal to 𝑒 prime 𝑣 plus 𝑒𝑣 prime. So to differentiate our function 𝑓 of π‘₯, we’ll set 𝑒 equal to 𝑒 to the power of π‘₯ and 𝑣 equal to the sec of π‘₯. Then, to find our function 𝑒 prime, we need to differentiate 𝑒 to the power of π‘₯. And we recall the derivative of 𝑒 to the power of π‘₯ with respect to π‘₯ is just equal to itself. So 𝑒 prime is equal to 𝑒 to the power of π‘₯.

Next, to find 𝑣 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, it’s useful to commit this to memory. Using this, we see that 𝑣 prime is equal to the sec of π‘₯ times the tan of π‘₯.

We’re now ready to apply the product rule to find the derivative of 𝑓 of π‘₯. It’s equal to 𝑒 prime 𝑣 plus 𝑒𝑣 prime. Substituting the functions we found for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime, we have that 𝑓 prime of π‘₯ is equal to 𝑒 to the power of π‘₯ times the sec of π‘₯ plus 𝑒 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 𝑒 to the π‘₯ and both terms share a factor of the sec of π‘₯.

So we take out our factor of 𝑒 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 𝑓 of π‘₯ is equal to 𝑒 to the power of π‘₯ times the sec of π‘₯, then the derivative of 𝑓 of π‘₯, 𝑓 prime of π‘₯, is equal to 𝑒 to the power of π‘₯ times the sec of π‘₯ multiplied by one plus the tan of π‘₯.

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