Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations Mathematics • Higher Education

Given that π₯ = 3π‘Β² + 1 and π¦ = 3π‘Β² + 5π‘, find dΒ²π¦/dπ₯Β².

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Video Transcript

Given that π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus five π‘, find the second derivative of π¦ with respect to π₯.

In this question, weβre asked to find the second derivative of π¦ with respect to π₯. And we can notice something interesting. π¦ is not given as a function in π₯. Instead, weβre given a pair of parametric equations in terms of the variable π‘. This means weβre going to need to differentiate this by using parametric differentiation.

And before we recall our formula for finding the second derivative of parametric equations, we can note that we know π₯ is a differentiable function in π‘ and π¦ is a differentiable function in π‘ because theyβre both polynomials. And this helps justify that we can use this formula to determine the second derivative of π¦ with respect to π₯, which we recall is equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by dπ₯ by dπ‘. And this is, of course, provided that dπ₯ by dπ‘ is nonzero.

To use this to determine the second derivative of π¦ with respect to π₯, we first need to find an expression for dπ¦ by dπ₯. And we can recall if π₯ and π¦ are given as a pair of parametric equations in terms of a variable π‘, then the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π‘ divided by the derivative of π₯ with respect to π‘. And once again, this is provided that π¦ and π₯ are differentiable functions in π‘ and dπ₯ by dπ‘ is not equal to zero.

Weβre now ready to start finding our expression for d two π¦ by dπ₯ squared. We see we first need to find expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs start by finding dπ¦ by dπ‘. This is equal to the derivative of three π‘ squared plus five π‘ with respect to π‘. Since this is a polynomial, we can do this term by term by using the power rule for differentiation. We multiply by the exponent of π‘ and then reduce this exponent by one. We get six π‘ plus five. We can follow the same process to determine dπ₯ by dπ‘. Thatβs the derivative of three π‘ squared plus one with respect to π‘. Once again, we apply the power rule for differentiation term by term. This time, we get six π‘.

We can now substitute these into our formula for dπ¦ by dπ₯. This then gives us that dπ¦ by dπ₯ is equal to six π‘ plus five divided by six π‘. And we could leave this expression like this. However, we know in our formula for d two π¦ by dπ₯ squared, weβre going to need to differentiate this with respect to π‘. And although we could evaluate this derivative by using the quotient rule, itβs easier to divide both terms in our numerator by six π‘. And since six π‘ divided by six π‘ is equal to one and five divided by six π‘ can be rewritten as five-sixths π‘ to the power of negative one, weβve shown dπ¦ by dπ₯ is equal to one plus five-sixths π‘ to the power of negative one.

And then we can just differentiate this by using the power rule for differentiation. And this is much easier to differentiate because we can just do this by using the power rule for differentiation. We could now start substituting our expressions into this formula. However, itβs easier to evaluate the numerator of the fraction on the right-hand side of the equation separately. We want to find the derivative of dπ¦ by dπ₯ with respect to π‘. Thatβs the derivative of five-sixths π‘ to the power of negative one with respect to π‘. And we can evaluate this term by term by using the power rule for differentiation. We get negative five-sixths π‘ to the power of negative two.

And we can simplify this slightly by using our laws of exponents. Multiplying by π‘ to the power of negative two is the same as dividing by π‘ squared. So we can rewrite this as negative five over six π‘ squared. Weβre now ready to substitute this and our expression for dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯ squared. This then gives us the second derivative of π¦ with respect to π₯ is equal to negative five over six π‘ squared all divided by six π‘. And we can then simplify this. Dividing by six π‘ is the same as multiplying by one over six π‘. And we can then evaluate this. Six π‘ squared times six π‘ is 36π‘ cubed. So this gives us negative five over 36π‘ cubed, which is our final answer.

Therefore, we were able to show if π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus five π‘, then the second derivative of π¦ with respect to π₯ is equal to negative five divided by 36π‘ cubed.