Video: Finding the Second Derivative of a Function Defined by Parametric Equations at a Given Value for the Parameter

If 𝑦 = βˆ’5π‘₯Β³ βˆ’ 7 and 𝑧 = 3π‘₯Β² + 16, find d²𝑧/d𝑦² at π‘₯ = 1.

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Video Transcript

If 𝑦 is equal to negative five π‘₯ cubed minus seven and 𝑧 is equal to three π‘₯ squared plus 16, find the second derivative of 𝑧 with respect to 𝑦 at π‘₯ is equal to one.

Since the question gives us 𝑦 as a function of π‘₯ and 𝑧 as a function of π‘₯, it might be tempting to try and write 𝑧 as a function of 𝑦. However, there’s a simpler method by using the chain rule.

First, we recall that the second derivative of 𝑧 with respect to 𝑦 is equal to the derivative with respect to 𝑦 of d𝑧 d𝑦. We recall that since both 𝑦 and 𝑧 are given as functions of π‘₯, we can calculate the derivative of 𝑧 with respect to 𝑦 by using the chain rule. As the derivative of 𝑧 with respect to π‘₯ multiplied by the derivative of π‘₯ with respect to 𝑦. And since we’re given 𝑦 as a function of π‘₯, instead of multiplying by the derivative of π‘₯ with respect to 𝑦, we can instead divide by the derivative of 𝑦 with respect to π‘₯.

We’re now ready to calculate these. We have the derivative of 𝑦 with respect to π‘₯ is equal to the derivative of negative five π‘₯ cubed minus seven with respect to π‘₯. To differentiate negative five π‘₯ cubed, we first multiply by the exponent, giving us negative five multiplied by three, which is negative 15. And then we reduce the exponent by one. And we know that the derivative of the constant negative seven is just equal to zero. So we have that d𝑦 dπ‘₯ is equal to negative 15π‘₯ squared.

Similarly, we have the derivative of 𝑧 with respect to π‘₯ is equal to the derivative of three π‘₯ squared plus 16 with respect to π‘₯. To differentiate three π‘₯ squared, first we multiply by the exponent. This gives us three multiplied by two, which is six. And then we reduce the exponent by one, giving us π‘₯ to the first power, which we can simplify to just be π‘₯. And the derivative of the constant 16 is just equal to zero. So we have the derivative of 𝑧 with respect to π‘₯ is equal to six π‘₯.

We’re now ready to use our chain rule formula. We have that d𝑧 d𝑦 is equal to d𝑧 dπ‘₯ divided by d𝑦 dπ‘₯. Substituting in d𝑧 dπ‘₯ is equal to six π‘₯ and d𝑦 dπ‘₯ is equal to negative 15π‘₯ squared gives us six π‘₯ divided by negative 15π‘₯ squared. We can simplify this by canceling the shared factor of π‘₯ in the numerator and the denominator and the shared factor of three in the numerator and the denominator. This gives us the derivative of 𝑧 with respect to 𝑦 is equal to negative two divided by five π‘₯.

We notice that our derivative function d𝑧 d𝑦 is a function of π‘₯. So let’s call this function 𝑓 of π‘₯. And we see that the question wants us to calculate the second derivative of 𝑧 with respect to 𝑦. That’s the same as taking the derivative of our function 𝑓 of π‘₯ with respect to 𝑦. And we already know how to calculate the derivative of a function of π‘₯ with respect to 𝑦 using the chain rule. This gives us the derivative of 𝑓 with respect to 𝑦 is equal to the derivative of 𝑓 with respect to π‘₯ multiplied by the derivative of π‘₯ with respect to 𝑦.

And just as we did before, instead of multiplying by the derivative of π‘₯ with respect to 𝑦, we can divide by the derivative of 𝑦 with respect to π‘₯. So what we have is the second derivative of 𝑧 with respect to 𝑦 is equal to the derivative with respect to 𝑦 of d𝑧 d𝑦, which we’ve defined to be equal to 𝑓 of π‘₯. And then using the chain rule, we have that this is equal to d𝑓 dπ‘₯ divided by d𝑦 dπ‘₯.

We have that the derivative of 𝑓 with respect to π‘₯ is equal to the derivative of negative two divided by five π‘₯ with respect to π‘₯. And we divide this by the derivative of 𝑦 with respect to π‘₯, which we found earlier to be equal to negative 15π‘₯ squared. We can rewrite negative two divided by five π‘₯ as negative two-fifths multiplied by π‘₯ to the power of negative one. And we can differentiate this as we did before. We multiply by the exponent of negative one, giving us two-fifths. And then we subtract one from the exponent, giving us π‘₯ to the power of negative two.

And we can simplify π‘₯ to the power of negative two by just dividing by π‘₯ squared, giving us two divided by five π‘₯ squared. And we can simplify division by negative 15π‘₯ squared to be equal to multiplying by negative one divided by 15π‘₯ squared. And if we simplify this, we’ve shown that the second derivative of 𝑧 with respect to 𝑦 is equal to negative two divided by 75π‘₯ to the fourth power. And we can think of this as a function of π‘₯.

If we go back to the question, we see that it wants us to find the second derivative of 𝑧 with respect to 𝑦 at the point where π‘₯ is equal to one. To find this, we want to substitute π‘₯ is equal to one into our equation for the second derivative of 𝑧 with respect to 𝑦. Doing this gives us that the second derivative of 𝑧 with respect to 𝑦 at the point π‘₯ is equal to one is equal to negative two divided by 75 multiplied by one to the fourth power. Which we can evaluate to be equal to negative two divided by 75.

Therefore, we’ve shown if 𝑦 is equal to negative five π‘₯ cubed minus seven and 𝑧 is equal to three π‘₯ squared plus 16. Then the second derivative of 𝑧 with respect to 𝑦 at π‘₯ is equal to one is equal to negative two divided by 75.

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