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If π₯ is equal to two multiplied by the sec of five π§ and the square root of three π¦ is equal to the tan of five π§, find the second derivative of π¦ with respect to π₯.
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Since the question asks us to find the second derivative of π¦ with respect to π₯, it might be tempting to try and write π¦ in terms of π₯.
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However, thereβs a simpler method by using the chain rule.
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We recall that calculating the second derivative of π¦ with respect to π₯ is the same as differentiating dπ¦ dπ₯ with respect to π₯.
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And we recall that the chain rule tells us 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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And instead of multiplying by the derivative of π§ with respect to π₯, since weβre given π₯ as a function of π§, we can do the equivalent, dividing by the derivative of π₯ with respect to π§.
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So letβs start by calculating the derivative of π₯ with respect to π§.
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Thatβs equal to the derivative of two multiplied by the sec of five π§ with respect to π§.
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And we recall that the derivative of the sec of ππ₯ with respect to π₯ is equal to π multiplied by the sec of ππ₯ multiplied by the tan of ππ₯.
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Applying this gives us that the derivative of π₯ with respect to π§ is equal to two multiplied by five, which we can simplify to 10, multiplied by the sec of five π§ multiplied by the tan of five π§.
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Calculating the derivative of π¦ with respect to π§ is more difficult since weβre not given π¦ directly as a function of π§.
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The simplest way to do this is to rearrange our equation to make π¦ the subject.
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Weβll do this by squaring both sides of the equation.
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Evaluating this gives us that three π¦ is equal to the tan squared of five π§.
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Next, we differentiate both sides of our equation with respect to π§.
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Differentiating three π¦ with respect to π§ gives us three multiplied by the derivative of π¦ with respect to π§.
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We can calculate the derivative of the tan of five π§ all squared with respect to π§.
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By recalling that, for a constant π, the derivative with respect to π₯ of the tan of π multiplied by π₯ is equal to π multiplied by the sec of ππ₯ squared.
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And by using the chain rule, we have that the derivative of the function π of π₯ all squared is equal to two multiplied by π prime of π₯ multiplied by π of π₯.
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Using this gives us that the derivative of the tan of five π§ all squared with respect to π§ is equal to two multiplied by our function π prime of π§.
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Which weβve shown to be equal to five multiplied by the sec of five π§ all squared.
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And then we multiply this by our function π of π§, which is the tan of five π§.
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We can simplify this by multiplying the two and the five to get 10 and then dividing both sides of our equation by three.
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This gives us that the derivative of π¦ with respect to π§ is equal to 10 over three multiplied by the sec of five π§ squared multiplied by the tan of five π§.
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Weβre now ready to apply our chain rule that the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π§ divided by the derivative of π₯ with respect to π§.
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We have a numerator of dπ¦ dπ§, which is equal to ten-thirds multiplied by the sec of five π§ all squared multiplied by the tan of five π§.
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And a denominator of dπ₯ by dπ§, which is equal to 10 multiplied by the sec of five π§ multiplied by the tan of five π§.
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We can then start canceling the shared factors.
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We can cancel the shared factor of the tan of five π§ in the numerator and the denominator.
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We can cancel one shared factor of the sec of five π§ in the numerator and the denominator.
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And we can cancel the shared factor of 10 in the numerator and the denominator.
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Therefore, weβve shown that the derivative of π¦ with respect to π₯ is equal to the sec of five π§ all divided by three.
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So weβve shown that our derivative function dπ¦ dπ₯ is a function of π§.
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Letβs call this π of π§.
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And the question wants us to calculate the second derivative of π¦ with respect to π₯.
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And thatβs the same as differentiating our function π of π§ with respect to π₯.
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And weβve already seen how to do this.
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We can use the chain rule.
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This gives us 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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And just as we did before, instead of multiplying by the derivative of π§ with respect to π₯, since weβre given π₯ as a function of π§, we can instead divide by the derivative of π₯ with respect to π§.
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So what we have is the second derivative of π¦ with respect to π₯ is equal to the derivative of dπ¦ dπ₯ with respect to π₯.
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And weβve called dπ¦ dπ₯ π of π§.
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And then using the chain rule, weβve shown that this is equal to dπ dπ§ divided by dπ₯ dπ§.
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Substituting in our expression for π, we get the derivative of π with respect to π§ is equal to the derivative of the sec of five π§ all divided by three with respect to π§.
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Then we can substitute in our expression for dπ₯ by dπ§.
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This is 10 sec of five π§ multiplied by the tan of five π§.
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We can differentiate the sec of five π§ to be equal to five multiplied by the sec of five π§ multiplied by the tan of five π§.
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And then we divide through by our constant of three.
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And instead of dividing, weβre going to multiply by the reciprocal.
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Simplifying this gives us five multiplied by the sec of five π§ multiplied by the tan of five π§ all divided by 30 multiplied by the sec of five π§ multiplied by the tan of five π§.
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We can then cancel the shared factor of the tan of five π§ in our numerator and our denominator and the shared factor of the sec of five π§ in the numerator and the denominator.
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And if we then cancel the shared factor of five, we get one-sixth.
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So weβve shown that if π₯ is equal to two multiplied by the sec of five π§ and the square root of three π¦ is equal to the tan of five π§, then the second derivative of π¦ with respect to π₯ is equal to one-sixth.