# Question Video: Finding the Second Derivative of Parametric Equations Mathematics • Higher Education

If π₯ = π^(βπ‘), π¦ = sin π‘, find dΒ²π¦/dπ₯Β².

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### Video Transcript

If π₯ is equal to π to the power of negative π‘, π¦ is equal to the sin of 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₯. However, 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 tells us we canβt just differentiate π¦ with respect to π₯ twice to determine d two π¦ by dπ₯ squared. Instead, weβre going to need to use parametric differentiation.

So letβs start by recalling how we find the second derivative of π¦ with respect to π₯ where π¦ is given as a function in π‘ and π₯ is given as a function in π‘. We can do this by using the following formula. Itβs equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by the derivative of π₯ with respect to π‘. And there is one thing worth noting about this. This formula uses dπ¦ by dπ₯. And we canβt directly find dπ¦ by dπ₯ since π¦ is a function in π‘ and π₯ is a function in π‘.

So weβre once again going to have to recall another formula for dπ¦ by dπ₯. Or we can derive this by using an application of the chain rule. We can also recall we can find dπ¦ by dπ₯ by finding dπ¦ by dπ‘ and dividing this by dπ₯ by dπ‘. And although expressions like dπ¦ by dπ₯, dπ¦ by dπ‘, and dπ₯ by dπ‘ are not fractions, we can treat them a little bit like fractions to help us remember certain results. For example, if we did treat them like fractions, we could think about multiplying by the reciprocal. Then, we could cancel the shared factor of dπ‘ in the numerator and denominator. This would then leave us with dπ¦ by dπ₯. And of course this isnβt exactly whatβs happening. However, it is a useful trick to help us remember this result.

Therefore, to find the second derivative of π¦ with respect to π₯, we need to start by finding dπ¦ by dπ₯. And to do this, we need to differentiate π¦ with respect to π‘ and π₯ with respect to π‘. So letβs start by finding the derivative of π¦ with respect to π‘. Thatβs equal to the derivative of the sin of π‘ with respect to π‘. We can do this by recalling the derivative of the sine function is just the cosine function. So this is just equal to the cos of π‘.

Letβs now find the derivative of π₯ with respect to π‘. Thatβs equal to the derivative of π to the power of negative π‘ with respect to π‘. We can do this by recalling for any real constant π, the derivative of π to the power of π times π with respect to π is π times π to the power of ππ. We just multiplied by the coefficient of the variable in the exponent. In this case, that coefficient is negative one. So we just get negative π to the power of negative π‘.

Now that weβve found dπ¦ by dπ‘ and dπ₯ by dπ‘, we can take that quotient to find dπ¦ by dπ₯. This then gives us the derivative of π¦ with respect to π₯ is equal to the cos of π‘ divided by negative π to the power of negative π‘. And we can simplify this by using our laws of exponents to bring π to the power of negative π‘ into the numerator. We get negative π to the power of π‘ times the cos of π‘. We could now substitute our expressions for dπ¦ by dπ₯ and dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯ squared. However, this would create a very complicated expression. So letβs instead evaluate the numerator first.

We need to differentiate our expression for dπ¦ by dπ₯ with respect to π‘. This means we need to find the derivative of negative π to the power of π‘ times the cos of π‘ with respect to π‘. This is the derivative of the product of two differentiable functions. So weβre going to need to use the product rule. This tells us if π of π‘ and π of π‘ are differentiable functions, then the derivative of π of π‘ multiplied by π of π‘ with respect to π‘ is equal to π prime of π‘ times π of π‘ plus π of π‘ multiplied by π prime of π‘. So to apply the product rule, weβre going to need to find expressions for the derivative of both of these factors.

Letβs start with the derivative of negative π to the power of π‘. π prime of π‘ will be equal to negative π to the power of π‘. Similarly, if we set π of π‘ to be the cos of π‘, then we know the derivative of the cosine function is negative the sine function. So we get π prime of π‘ is negative sin π‘. Substituting these expressions into the product rule, we get negative π to the power of π‘ times the cos of π‘ plus negative π to the power of π‘ multiplied by negative the sin of π‘. And we can simplify this expression. In our second term, negative one times negative one is just equal to one. This leaves us with negative π to the power of π‘ times the cos of π‘ plus π to the power of π‘ times the sin of π‘. We can also take our shared factor of π to the power of π‘. This then leaves us with π to the power of π‘ multiplied by the sin of π‘ minus the cos of π‘.

Weβre not done yet, however. Remember, we still need to divide this by dπ₯ by dπ‘ to determine the second derivative of π¦ with respect to π₯. So letβs clear some space and substitute these expressions into our formula. We get the second derivative of π¦ with respect to π₯ is equal to π to the power of π‘ multiplied by the sin of π‘ minus the cos of π‘ all divided by negative π to the power of negative π‘. And we can simplify this expression. First, dividing by π to the power of negative π‘ is the same as multiplying by π to the power of π‘. This gives us two factors of π to the power of π‘ in our numerator. We can simplify this to give us π to the power of two π‘. And taking the factor of negative one into our numerator, we get negative π to the power of two π‘ multiplied by the sin of π‘ minus the cos of π‘.

Now, all thatβs left to do is distribute the negative over our parentheses and then reorder the two terms in the parentheses. This then gives us our final answer. Therefore, weβve shown if π₯ is equal to π to the power of negative π‘ and π¦ is equal to the sin of π‘, then the second derivative of π¦ with respect to π₯ is equal to π to the power of two π‘ multiplied by the cos of π‘ minus the sin of π‘.

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