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

By using parametric differentiation, determine the derivative of 5𝑥³ + 𝑥² − 2 with respect to 4𝑥² + 8.

02:51

### Video Transcript

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

We recall that if we had 𝑦 is equal to some function 𝑓 of 𝑥 and 𝑧 is equal to some function 𝑔 of 𝑥, then by the chain rule. The derivative of 𝑦 with respect to 𝑧 is equal to the derivative of 𝑦 with respect to 𝑥 multiplied by the derivative of 𝑥 with respect to 𝑧. Then by setting 𝑥 to be equal to the inverse of 𝑔 of 𝑧, we have that the derivative of 𝑥 with respect to 𝑧 is equal to one divided by the derivative of 𝑧 with respect to 𝑥 by using our inverse function theorem. Using this, we can calculate the derivative of 𝑦 with respect to 𝑧 by first calculating the derivative of 𝑦 with respect to 𝑥 and then dividing this by the derivative of 𝑧 with respect to 𝑥.

The question wants us to calculate the derivative of five 𝑥 cubed plus 𝑥 squared minus two. And it wants us to do this with respect to four 𝑥 squared plus eight. So if we set 𝑦 equal to five 𝑥 cubed plus 𝑥 squared minus two and 𝑧 equal to four 𝑥 squared plus eight, then d𝑦 d𝑧 is what the question wants us to calculate. It’s the derivative of five 𝑥 cubed plus 𝑥 squared minus two with respect to four 𝑥 squared plus eight. And we have that this is equal to the derivative of 𝑦 with respect to 𝑥 divided by the derivative of 𝑧 with respect to 𝑥.

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

Finally, the derivative of any constant is just equal to zero. So the derivative of negative two is just equal to zero. So we’ve shown that d𝑦 d𝑥 is equal to 15𝑥 squared plus two 𝑥. We can do the same to calculate the derivative of 𝑧 with respect to 𝑥. That’s the derivative of four 𝑥 squared plus eight with respect to 𝑥.

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

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