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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₯.
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The question gives us a pair of parametric equations and asks us to find the second derivative of π¦ with respect to π₯.
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So we might be tempted to first try writing these into a Cartesian form.
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However, this is not necessary as we can use the chain rule.
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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₯.
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This means we can use the chain rule to find our first derivative of π¦ with respect to π₯.
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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‘.
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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.
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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‘.
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We know that the derivative of π‘ with respect to π‘ is just equal to one.
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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‘.
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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‘.
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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‘.
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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‘.
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Since weβve now found an expression for dπ¦ dπ‘, according to the chain rule, we should be now finding an expression for dπ‘ dπ₯.
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However, it will be difficult to directly find a value for dπ‘ dπ₯ since π₯ is a function of π‘.
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We would need to rearrange our equation for π₯ to give π‘ as a function of π₯.
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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₯.
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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‘.
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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₯.
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Weβre now ready to use our chain rule, which says that dπ¦ dπ₯ is equal to dπ¦ dπ‘ multiplied by dπ‘ dπ₯.
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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‘.
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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‘.
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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‘.
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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‘.
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But the question is asking us to find the second derivative of π¦ with respect to π₯.
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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₯.
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However, we know that the derivative of π¦ with respect to π₯ is a function of π‘.
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This means itβll be difficult to differentiate this directly with respect to π₯.
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So weβre going to have to use the chain rule again.
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So we go back to our chain rule.
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But this time, instead of differentiating π¦, weβre going to differentiate π, which is equal to the derivative of π¦ with respect to π₯.
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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₯.
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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₯.
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We have that π is a function of π‘.
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So we can calculate the derivative of π with respect to π‘.
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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‘.
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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‘.
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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₯.
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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.
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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.
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And the derivative of π£ with respect to π‘ is equal to 16 multiplied by π to the power of four π‘.
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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.
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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‘.
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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‘.
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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‘.
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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.
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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‘.
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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‘.