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Given that π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus five π‘, find d two π¦ by dπ₯ squared.
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In this question, weβve been given a pair of parametric equations.
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And weβve been asked to find d two π¦ by dπ₯ squared, which is the second derivative of π¦ with respect to π₯.
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Now, we in fact have an equation to help us find the second derivative of parametric equations.
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The equation tells us that d two π¦ by dπ₯ squared is equal to d by dπ‘ of dπ¦ by dπ₯ over dπ₯ by dπ‘.
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In order to use this equation, we also need to find dπ¦ by dπ₯.
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And we also have an equation for finding dπ¦ by dπ₯ when given parametric equations.
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This equation tells us that dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘.
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We can start our solution by finding dπ¦ by dπ‘ and dπ₯ by dπ‘.
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We have that π₯ is equal to three π‘ squared plus one.
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Using the power rule for differentiation, we multiply by the power and decrease the power by one.
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And so, differentiating the term three π‘ squared, we obtain six π‘.
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Since one is a constant and we differentiate it, it will go to zero.
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Therefore, we have that dπ₯ by dπ‘ is equal to six π‘.
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Weβve also been given that π¦ is equal to three π‘ squared plus five π‘.
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This can again be differentiated using the power rule.
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Differentiating the first term, we again get six π‘.
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And when we differentiate the five π‘, we simply get five.
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Therefore, we have that dπ¦ by dπ‘ is equal to six π‘ plus five.
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Weβve now found all the components in order to find dπ¦ by dπ₯.
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We obtain that dπ¦ by dπ₯ is equal to six π‘ plus five over six π‘.
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When we look at our formula for the second derivative of π¦ with respect to π₯, we spot that we have to find d by dπ‘ of dπ¦ by dπ₯.
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So thatβs d by dπ‘ of six π‘ plus five over six π‘, which is a quotient.
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Therefore, we can use the quotient rule to help us differentiate here.
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We find that the derivative of a quotient of some functions, π’ over π£, is equal to π£ multiplied by dπ’ by dπ₯ minus π’ multiplied by dπ£ by dπ₯ all over π£ squared.
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Now, in our case, our numerator is equal to six π‘ plus five.
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Therefore, itβs equal to π’.
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And our denominator is six π‘.
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And so itβs equal to π£.
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We can differentiate six π‘ plus five with respect to π‘ to find that dπ’ by dπ‘ is equal to six.
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And differentiating six π‘ with respect to π‘, we find that dπ£ by dπ‘ is also equal to six.
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Now, weβre ready to substitute π’, π£, dπ’ by dπ‘, and dπ£ by dπ‘ into the formula given to us by the quotient rule.
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What we get is six π‘ times six minus six π‘ plus five times six all over six π‘ squared.
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Multiplying through, we obtain 36π‘ minus 36π‘ minus 30 over 36π‘ squared.
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Therefore, we can cancel the 36π‘ with the minus 36π‘.
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And we can cancel through our factor of five to obtain that d by dπ‘ of dπ¦ by dπ₯ is equal to negative five over six π‘ squared.
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So if we look back at our formula for d two π¦ by dπ₯ squared, we have now found d by dπ‘ of dπ¦ by dπ₯.
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And itβs equal to negative five over six π‘ squared.
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We also found dπ₯ by dπ‘ earlier.
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And itβs equal to six π‘.
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Using our formula, we have that d two π¦ by dπ₯ squared is equal to negative five over six π‘ squared over six π‘, which simplifies to give us a solution that d two π¦ by dπ₯ squared is equal to negative five over 36π‘ cubed.