Video: Differentiating the Quotient of Functions Involving Exponential and Trigonometric Functions

Find the first derivative of the function 𝑦 = 𝑒^(5π‘₯)/(7 tan 8π‘₯).

04:14

Video Transcript

Find the first derivative of the function 𝑦 is equal to 𝑒 to the power of five π‘₯ divided by seven times the tan of eight π‘₯.

We’re given a function 𝑦 which is the quotient of two functions. And we need to find the first derivative of 𝑦. Since 𝑦 is a function of π‘₯, this will be the first derivative of 𝑦 with respect to π‘₯.

There’s a few different ways we could approach this. For example, we might be tempted to try using our trigonometric identities to rewrite this into an easier function to differentiate. Using these, we could rewrite 𝑦 as 𝑒 to the power of five π‘₯ multiplied by the cot of eight π‘₯ all divided by seven. And of course, we could evaluate the derivative of this expression with respect to π‘₯ by using the product rule.

However, that’s not the only method we could use to approach this problem. We could also just differentiate this by using the quotient rule. Either method is valid. And we can use whichever one we prefer. In this video, we’re going to use the quotient rule.

We recall the quotient rule tells us the derivative of the quotient of two differentiable functions 𝑒 over 𝑣 is equal to 𝑒 prime 𝑣 minus 𝑣 prime 𝑒 all divided by 𝑣 squared. And since we can’t divide by zero, this will only be valid when our denominator of 𝑣 squared is not equal to zero. So to apply the quotient rule, we’ll set 𝑒 of π‘₯ to be the function in our numerator, that’s 𝑒 to the power of five π‘₯, and 𝑣 of π‘₯ to be the function in our denominator, that’s seven tan of eight π‘₯.

Now, to apply the quotient rule, we’re 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 five π‘₯ with respect to π‘₯. To evaluate this derivative, we need to recall one of our standard rules for differentiating exponential functions. For any real constant π‘Ž, the derivative of 𝑒 to the power of π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times 𝑒 to the power of π‘Žπ‘₯.

So to differentiate 𝑒 of π‘₯, we can see our value of π‘Ž is equal to five. So we set π‘Ž equal to five, and we get 𝑒 prime of π‘₯ is equal to five times 𝑒 to the power of five π‘₯. Next, we need to find an expression for 𝑣 prime of π‘₯. That’s the derivative of seven times the tan of eight π‘₯ with respect to π‘₯. And to evaluate this derivative, we need to recall one of our standard trigonometric derivative results. For any real constant π‘Ž, the derivative of the tan of π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times the sec squared of π‘Žπ‘₯.

This time, the value of π‘Ž, which is the coefficient of π‘₯, is equal to eight. So the derivative of the tan of eight π‘₯ with respect to π‘₯ will be eight times the sec squared of eight π‘₯. Of course, we need to multiply this by seven to find 𝑣 prime of π‘₯. And in this expression, we can simplify seven multiplied by eight to give us 56.

We’re now ready to find an expression for the derivative of 𝑦 with respect to π‘₯ by using the quotient rule. It’s equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ minus 𝑣 prime of π‘₯ times 𝑒 of π‘₯ all divided by 𝑣 of π‘₯ all squared. Substituting in our expressions for 𝑒 of π‘₯, 𝑒 prime of π‘₯, 𝑣 of π‘₯, and 𝑣 prime of π‘₯, we get d𝑦 by dπ‘₯ is equal to five times 𝑒 to the power of five π‘₯ multiplied by seven tan of eight π‘₯ minus 56 times the sec squared of eight π‘₯ multiplied by 𝑒 to the power of five π‘₯ all divided by seven tan of eight π‘₯ all squared.

And we can simplify this expression by taking out the shared factor of 𝑒 to the power of five π‘₯ in our numerator. This gives us 𝑒 to the power of five π‘₯ multiplied by 35 tan of eight π‘₯ minus 56 sec squared of eight π‘₯ all divided by seven tan of eight π‘₯ all squared.

And we could leave our answer like this. However, we can also simplify this even more. We’ll start by distributing the exponent in our denominator. This will give us a new denominator of seven squared multiplied by the tan squared of eight π‘₯. And of course, seven squared is equal to 49. And now, we can see our numerator and our denominator all share a factor of seven. So we’ll divide both our numerator and our denominator through by seven. And this gives us our final answer.

Therefore, given the function 𝑦 is equal to 𝑒 to the power of five π‘₯ divided by seven times the tan of eight π‘₯, we were able to show the first derivative of 𝑦 with respect to π‘₯ is equal to 𝑒 to the power of five π‘₯ multiplied by five times the tan of eight π‘₯ minus eight times the sec squared of eight π‘₯ all divided by seven times the tan squared of eight π‘₯.

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