### Video Transcript

Given that π₯ is equal to two multiplied by π to the power of two π‘ and that π¦ is equal to π‘ multiplied by π to the power of negative two π‘, find the second derivative of π¦ with respect to π₯.

The question gives us a pair of parametric equations and asks us to find the second derivative of π¦ with respect to π₯. So we might be tempted to first try writing these into a Cartesian form. However, this is not necessary as we can use the chain rule.

We recall that the chain rule states that we can calculate the derivative of π¦ with respect to π₯ by first differentiating π¦ with respect to π‘ and then multiplying this by the derivative of π‘ with respect to π₯. This means we can use the chain rule to find our first derivative of π¦ with respect to π₯.

We start by finding the derivative of π¦ with respect to π‘, which is equal to the derivative of π‘ multiplied by π to the negative two π‘ with respect to π‘. Weβll recall that the product rule for differentiation says that the derivative of the product of two functions, π and π, with respect to π‘ is equal to the derivative of π with respect to π‘ multiplied by π plus the derivative of π with respect to π‘ multiplied by π. This gives us that the derivative of π‘ multiplied by π to the negative two π‘ with respect to π‘ is equal to the derivative of π‘ with respect to π‘ multiplied by π to the negative two π‘ plus the derivative of π to the negative two π‘ multiplied by π‘.

We know that the derivative of π‘ with respect to π‘ is just equal to one. And we also know that the derivative with respect to π‘ of π to the power of ππ‘ is just equal to π multiplied by π to the power of ππ‘. So our derivative of π to the negative two π‘ is just equal to negative two multiplied by π to the power of negative two π‘. Giving us that dπ¦ dπ‘ is equal to π to the negative two π‘ minus two π‘ multiplied by π to the negative two π‘.

And at this point, we could also take out the factor of π to the power of negative two π‘ to give us π to the power of negative two π‘ multiplied by one minus two π‘. Since weβve now found an expression for dπ¦ dπ‘, according to the chain rule, we should be now finding an expression for dπ‘ dπ₯.

However, it will be difficult to directly find a value for dπ‘ dπ₯ since π₯ is a function of π‘. We would need to rearrange our equation for π₯ to give π‘ as a function of π₯. However, we can use the fact that dπ‘ dπ₯ is the reciprocal of dπ₯ dπ‘ to first find the derivative of π₯ with respect to π‘ and then use this to find the derivative of π‘ with respect to π₯.

We have that the derivative of π₯ with respect to π‘ is equal to the derivative of two multiplied by π to the power of two π‘ with respect to π‘, which we can evaluate to give us four multiplied by π to the power of two π‘. We then take the reciprocal of this to see that one divided by four multiplied by π to the power of two π‘ is equal to the derivative of π‘ with respect to π₯.

Weβre now ready to use our chain rule, which says that dπ¦ dπ₯ is equal to dπ¦ dπ‘ multiplied by dπ‘ dπ₯. We showed earlier that the derivative of π¦ with respect to π‘ is equal to π to the negative two π‘ multiplied by one minus two π‘. And we also showed that the derivative of π‘ with respect to π₯ is equal to one divided by four multiplied by π to the power of two π‘. If we bring our π to the negative two π‘ turned down into the denominator, we will get the derivative of π¦ with respect to π₯ is equal to one minus two π‘ divided by π to the power of two π‘ multiplied by four π to the two π‘.

And finally, we can simplify this to give us that dπ¦ dπ₯ is equal to one minus two π‘ divided by four multiplied by π to the power of four π‘. But the question is asking us to find the second derivative of π¦ with respect to π₯. We know that the second derivative of π¦ with respect to π₯ is just equal to the derivative of the derivative of π¦ with respect to π₯. However, we know that the derivative of π¦ with respect to π₯ is a function of π‘. This means itβll be difficult to differentiate this directly with respect to π₯. So weβre going to have to use the chain rule again.

So we go back to our chain rule. But this time, instead of differentiating π¦, weβre going to differentiate π, which is equal to the derivative of π¦ with respect to π₯. We see that the derivative of π with respect to π₯ is equal to the derivative of π with respect to π‘ multiplied by the derivative of π‘ with respect to π₯. Since we chose our function π to be the derivative of π¦ with respect to π₯, the derivative of π with respect to π₯ is our second derivative of π¦ with respect to π₯.

We have that π is a function of π‘. So we can calculate the derivative of π with respect to π‘. And we calculated earlier that the derivative of π‘ with respect to π₯ is equal to one divided by four multiplied by π to the power of two π‘. What this means is the second derivative of π¦ with respect to π₯ is equal to the derivative of one minus two π‘ over four multiplied by π to the four π‘ with respect to π‘ multiplied by one divided by four multiplied by π to the two π‘. And this is just an application of our chain rule where, instead of differentiating π¦, weβre differentiating the function π, which weβve defined to equal the derivative of π¦ with respect to π₯.

We can now use the quotient rule, which says that the derivative of the quotient of π’ and π£ is equal to the derivative of π’ with respect to π‘ multiplied by π£ minus the derivative of π£ with respect to π‘ multiplied by π’ all divided by π£ squared. This means if we let π’ equal one minus two π‘ and π£ equal four multiplied by π to the power of four π‘, then we have the derivative of π’ with respect to π‘ is equal to negative two. And the derivative of π£ with respect to π‘ is equal to 16 multiplied by π to the power of four π‘. Giving us that the derivative of the quotient of π’ and π£ is equal to negative two multiplied by four π to the four π‘ minus 16 multiplied by π to the four π‘ multiplied by one minus two π‘ all divided by four π to the four π‘ all squared.

We can then multiply out the parentheses in the numerator and to evaluate the square in the denominator to get negative eight multiplied by π to the four π‘ minus 16π to the four π‘ plus 32π‘π to the four π‘ all divided by 16π to the eight π‘. We can then combine the like terms in the numerator and then rearrange to get a new numerator of 32 multiplied by π‘π to the four π‘ minus 24π to the power of four π‘.

We can then cancel the shared factor of eight multiplied by π to the four π‘ in both the numerator and the denominator, giving us four π‘ minus three divided by two multiplied by π to the power of four π‘. We then substitute this back into our formula for the second derivative of π¦ with respect to π₯, which we found using the chain rule. Giving us that the second derivative of π¦ with respect to π₯ is equal to four π‘ minus three divided by two π to the four π‘ all multiplied by one divided by four π to the two π‘. Which we can then calculate to give us that the second derivative of π¦ with respect to π₯ is equal to four π‘ minus three divided by eight multiplied by π to the power of six π‘.