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

By using parametric differentiation, determine the derivative of 5π‘₯Β³ + π‘₯Β² βˆ’ 2 with respect to 4π‘₯Β² + 8.

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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 𝑦 is equal to some function 𝑓 of π‘₯ and 𝑧 is equal to some function 𝑔 of π‘₯, then by the chain rule. The derivative of 𝑦 with respect to 𝑧 is equal to the derivative of 𝑦 with respect to π‘₯ multiplied by the derivative of π‘₯ with respect to 𝑧. Then by setting π‘₯ to be equal to the inverse of 𝑔 of 𝑧, we have that the derivative of π‘₯ with respect to 𝑧 is equal to one divided by the derivative of 𝑧 with respect to π‘₯ 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 π‘₯ and then dividing this by the derivative of 𝑧 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 𝑦 equal to five π‘₯ cubed plus π‘₯ squared minus two and 𝑧 equal to four π‘₯ squared plus eight, then d𝑦 d𝑧 is what the question wants us to calculate. It’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 𝑦 with respect to π‘₯ divided by the derivative of 𝑧 with respect to π‘₯.

We can now calculate both of these. The derivative of 𝑦 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 we’ve shown that d𝑦 dπ‘₯ is equal to 15π‘₯ squared plus two π‘₯. We can do the same to calculate the derivative of 𝑧 with respect to π‘₯. That’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 𝑧 with respect to π‘₯ is equal to eight π‘₯. Substituting these into our equation gives us that the derivative of 𝑦 with respect to 𝑧 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‘₯.

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