# Video: Differentiating a Function with Respect to Another Function Using Parametric Differentiation

By using parametric differentiation, determine the derivative of 5π₯Β³ + π₯Β² β 2 with respect to 4π₯Β² + 8.

02:51

### Video Transcript

By using parametric differentiation, determine the derivative of five π₯ cubed plus π₯ squared minus two with respect to four π₯ squared plus eight.

We recall that if we had π¦ is equal to some function π of π₯ and π§ is equal to some function π of π₯, then by the chain rule. The derivative of π¦ with respect to π§ is equal to the derivative of π¦ with respect to π₯ multiplied by the derivative of π₯ with respect to π§. Then by setting π₯ to be equal to the inverse of π of π§, we have that the derivative of π₯ with respect to π§ is equal to one divided by the derivative of π§ with respect to π₯ by using our inverse function theorem. Using this, we can calculate the derivative of π¦ with respect to π§ by first calculating the derivative of π¦ with respect to π₯ and then dividing this by the derivative of π§ with respect to π₯.

The question wants us to calculate the derivative of five π₯ cubed plus π₯ squared minus two. And it wants us to do this with respect to four π₯ squared plus eight. So if we set π¦ equal to five π₯ cubed plus π₯ squared minus two and π§ equal to four π₯ squared plus eight, then dπ¦ dπ§ is what the question wants us to calculate. Itβs the derivative of five π₯ cubed plus π₯ squared minus two with respect to four π₯ squared plus eight. And we have that this is equal to the derivative of π¦ with respect to π₯ divided by the derivative of π§ with respect to π₯.

We can now calculate both of these. The derivative of π¦ with respect to π₯ is equal to the derivative of five π₯ cubed plus π₯ squared minus two with respect to π₯. We can differentiate this term by term. To differentiate five π₯ cubed, we multiply it by the exponent and then reduce the exponent by one. This gives us 15π₯ squared. We do the same to differentiate π₯ squared. We multiply by the exponent of two and then reduce the exponent by one, giving us two π₯.

Finally, the derivative of any constant is just equal to zero. So the derivative of negative two is just equal to zero. So weβve shown that dπ¦ dπ₯ is equal to 15π₯ squared plus two π₯. We can do the same to calculate the derivative of π§ with respect to π₯. Thatβs the derivative of four π₯ squared plus eight with respect to π₯.

To differentiate four π₯ squared, we multiply by the exponent of two and then reduce the exponent by one, which gives us eight π₯. And then to differentiate the constant eight, we just get zero, giving us that the derivative of π§ with respect to π₯ is equal to eight π₯. Substituting these into our equation gives us that the derivative of π¦ with respect to π§ is equal to 15π₯ squared plus two π₯ all divided by eight π₯.

Therefore, we have shown that the derivative of five π₯ cubed plus π₯ squared minus two with respect to four π₯ squared plus eight is equal to 15π₯ squared plus two π₯ all divided by eight π₯.