Video: Finding the First Derivative of the Quotient of Trigonometric and Linear Functions Using the Quotient Rule

If 𝑦 = (tan 7π‘₯)/(9π‘₯), find d𝑦/dπ‘₯.

02:39

Video Transcript

If 𝑦 is equal to the tan of seven π‘₯ divided by nine π‘₯, find the derivative of 𝑦 with respect to π‘₯.

We’re given 𝑦 as the quotient of two functions. It’s the tan of seven π‘₯ divided by nine π‘₯. We need to use this to find the derivative of 𝑦 with respect to π‘₯. And there’s a few different ways of doing this. For example, we could consider this as the tan of seven π‘₯ multiplied by one over nine π‘₯ and then use the product rule.

However, since 𝑦 is given as the quotient of two functions, we’ll do this by using the quotient rule. First, we need to recall the quotient rule. It tells us the derivative of the quotient of two functions 𝑒 over 𝑣 is equal to 𝑒 prime 𝑣 minus 𝑣 prime 𝑒 all divided by 𝑣 squared.

So to apply the quotient rule to 𝑦, we need to set 𝑒 of π‘₯ to be the function in our numerator β€” that’s the tan of seven π‘₯ β€” and 𝑣 of π‘₯ to be the function in our denominator β€” that’s nine π‘₯. We then need to find expressions for 𝑒 prime and 𝑣 prime.

Let’s start with 𝑒 prime of π‘₯. That’s the derivative of the tan of seven π‘₯ with respect to π‘₯. And to evaluate this, we need to recall one of our standard trigonometric derivative results. We know 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‘₯. In our case, the value of π‘Ž is equal to seven. So using this, we’ve shown 𝑒 prime of π‘₯ is equal to seven times the sec squared of seven π‘₯.

We can also find an expression for 𝑣 prime of π‘₯. It’s the derivative of nine π‘₯ with respect to π‘₯. But nine π‘₯ is a linear function. So its slope will just be the coefficient of π‘₯, which is nine.

We’re now ready to use the quotient rule to find an expression for d𝑦 by dπ‘₯. 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‘₯, 𝑣 of π‘₯, 𝑒 prime of π‘₯, and 𝑣 prime of π‘₯, we get that d𝑦 by dπ‘₯ is equal to seven times the sec squared of seven π‘₯ multiplied by nine π‘₯ minus nine times the tan of seven π‘₯ all divided by nine π‘₯ all squared.

And now we can start simplifying. First, in our denominator, nine π‘₯ all squared can be simplified to give us 81π‘₯ squared. Next, we can cancel a shared factor of nine in our numerator and our denominator. And then, after some rearrangements, we get seven π‘₯ times the sec squared of seven π‘₯ minus the tan of seven π‘₯ all divided by nine π‘₯ squared. And this is our final answer.

Therefore, by using the quotient rule on 𝑦 is equal to the tan of seven π‘₯ all divided by nine π‘₯, we were able to show that d𝑦 by dπ‘₯ is equal to seven π‘₯ times the sec squared of seven π‘₯ minus the tan of seven π‘₯ all divided by nine π‘₯ squared.

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