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If π₯ is equal to the cot of π‘ and π¦ is equal to the csc of π‘, find the second derivative of π¦ with respect to π₯.
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In this question, weβre asked to determine the second derivative of π¦ with respect to π₯.
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However, weβre not given π¦ as a function in π₯.
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Instead, weβre given a pair of parametric equations in terms of the variable π‘.
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And since π¦ is not given as a function in π₯, we canβt just differentiate our expression for π¦ with respect to π₯ twice to determine dπ¦ by dπ₯ squared.
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Instead, weβre going to need to use parametric differentiation.
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We can then recall that d two π¦ by dπ₯ squared will be equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by the derivative of π₯ with respect to π‘.
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And there is one small thing worth noting about this formula.
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We need to use dπ¦ by dπ₯.
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However, π¦ is not giving as a function of π₯, so we canβt just differentiate π¦ with respect to π₯.
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And we can find this by once again using parametric differentiation.
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Or alternatively, itβs an application of the chain rule. dπ¦ by dπ₯ will be equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘.
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Therefore, to find the second derivative of π¦ with respect to π₯, weβre going to first need to find expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘.
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Letβs start by finding the derivative of π¦ with respect to π‘.
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Thatβs the derivative of the csc of π‘ with respect to π‘.
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And we can do this by recalling the derivative of the csc of π with respect to π is equal to negative the cot of π multiplied by the csc of π.
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Therefore, the derivative of π¦ with respect to π‘ is equal to negative the cot of π‘ multiplied by the csc of π‘.
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We can follow a very similar process to determine the derivative of π₯ with respect to π‘.
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Thatβs the derivative of the cot of π‘ with respect to π‘.
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In this case, we recall the derivative of the cot of π with respect to π is equal to negative the csc squared of π.
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Therefore, dπ₯ by dπ‘ is equal to negative the csc squared of π‘.
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We can now substitute our expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘ into our formula to determine dπ¦ by dπ₯.
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Doing this, we get that dπ¦ by dπ₯ is equal to negative the cot of π‘ times the csc of π‘ divided by negative the csc squared of π‘.
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And we can simplify this.
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We can start by canceling the shared factor of negative one in the numerator and denominator.
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And we can also cancel one shared factor of csc of π‘ in the numerator and denominator.
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This leaves us with the cot of π‘ divided by the csc of π‘.
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Remember, to find d two π¦ by dπ₯ squared, weβre going to need to differentiate dπ¦ by dπ₯ with respect to π‘.
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So itβs a good idea to try and make this expression as simple as possible.
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So while we could leave this as cot of π‘ divided by the csc of π‘ and differentiate this by using the quotient rule, letβs simplify this by using our trigonometric identities.
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Weβre going to use the fact that the csc of π‘ is equal to the cos of π‘ divided by the sin of π‘.
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And one over the csc of π‘ is equal to the sin of π‘.
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Therefore, we can rewrite our numerator as the cos of π‘ divided by the sin of π‘.
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And weβre multiplying this by one over the csc of π‘.
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We can instead multiply it by the sin of π‘.
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This gives us the cos of π‘ divided by the sin of π‘ multiplied by the sin of π‘.
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And we can simplify this by canceling the shared factor of sin of π‘ in the numerator and denominator.
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Therefore, dπ¦ by dπ₯ is equal to the cos of π‘.
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Weβre now ready to use our formula to find an expression for the second derivative of π¦ with respect to π₯.
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Letβs start by clearing some space.
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Substituting in our expression for dπ¦ by dπ₯ and our expression for dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯ squared, we get the second derivative of π¦ with respect to π₯ is equal to the derivative of the cos of π‘ with respect to π‘ divided by negative the csc squared of π‘.
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Letβs start by evaluating the numerator.
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We recall the derivative of the cos of π with respect to π is negative the sin of π.
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So the derivative of the cos of π‘ with respect to π‘ is negative the sin of π‘.
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We can also recall that dividing by the csc of π‘ is the same as multiplying by the sin of π‘.
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So dividing by the csc squared of π‘ is the same as multiplying by the sin squared of π‘.
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We can also move the negative one from our denominator into our numerator.
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However, this will just cancel with our other factor of negative one.
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This just leaves us with the sin of π‘ times the sin squared of π‘, which is of course equal to the sin cubed of π‘, which is our final answer.
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Therefore, we were able to show if π₯ is the cot of π‘ and π¦ is the csc of π‘, then the second derivative of π¦ with respect to π₯ is the sin cubed of π‘.