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If sin of π¦ plus cos of two π₯ equals zero, prove that d two π¦ by dπ₯ squared minus dπ¦ by dπ₯ squared times tan of π¦ is equal to four cos of two π₯ times sec of π¦.
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In order to start answering this question, letβs look at what weβre trying to prove.
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And we can see that we have a d two π¦ by dπ₯ squared term and a dπ¦ by dπ₯ term.
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And these tell us that there must be a differential with respect to π₯.
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And that d two π¦ by dπ₯ squared term is a second-order differential, which tells us that something has been differentiated with respect to π₯ twice.
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Using this information that weβve just found from what weβre trying to prove, the next logical step will be to differentiate the equation given in the question with respect to π₯ twice.
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In order to do this, letβs quickly recall the chain rule.
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The chain rule tells us that dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ but dπ₯, where π’ is another variable.
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And this makes it easier to differentiate a function within a function.
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Now letβs try to differentiate our equation.
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Now since there are terms which are differentiated with respect to π₯ in the thing weβre trying to prove, weβll differentiate with respect to π₯ here.
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And weβre left with d by dπ₯ of sin of π¦ plus cos of two π₯ is equal to d by dπ₯ of zero.
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Now this type of differentiation where we just differentiate a whole equation is called implicit differentiation.
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To start off here, we can evaluate the right-hand side, since if you differentiate a constant by anything it will give you zero.
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So d by π₯ of zero is simply zero.
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And now for the left-hand side of the equation, we can split the terms into two differentiations, since addition within differentiation can be split up like this. d by dπ₯ of sin of π¦ plus d by dπ₯ of cos of two π₯ is equal to zero.
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And now in order to resolve the d by dπ₯ of cos of two π₯, we can use the chain rule.
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If we substitute in π’ is equal to two π₯, then the chain rule tells us that this differential becomes d by dπ’ of cos of π’ times dπ’ by dπ₯.
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Then cos of π’ differentiates to negative sin of π’.
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And we still need to multiply by the dπ’ by dπ₯.
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And now we can substitute back in π’ is equal to two π₯, which gives us negative sin of two π₯ times d by dπ₯ of two π₯.
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And in order to differentiate two π₯ with respect to π₯, we simply multiply by the power, which is one, and then decrease the power by one.
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And so it becomes two times π₯ to the power of zero, which is simply two.
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And so now we found the differential of cos of two π₯ with respect to π₯.
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And it is negative two sin of two π₯.
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And now in order to differentiate sin of π¦ with respect to π₯, we will again be using the chain rule.
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However, we can adapt the chain rule slightly to make this easier for ourselves.
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If we substitute π¦ is equal to π of π¦ into the equation for the chain rule, we end up with d by dπ₯ of π of π¦ is equal to d by dπ’ of π of π¦ times dπ’ by dπ₯.
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However, weβre allowed to choose our variable π’.
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And so we can let π’ be equal to π¦.
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And so we end up with d by dπ₯ of π of π¦ is equal to d by dπ¦ of π of π¦ timesed by dπ¦ by dπ₯.
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And so we can apply this to the function weβre trying to resolve.
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So thatβs d by dπ₯ of sin of π¦.
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Now in our case, π of π¦ is equal to sin of π¦.
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This gives us that d by dπ₯ of sin of π¦ is equal to d by dπ¦ of sin of π¦ times dπ¦ by dπ₯.
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Then d by dπ¦ of sin of π¦ is simply cos of π¦.
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And so we get that d by dπ₯ of sin of π¦ is equal to cos of π¦ times dπ¦ by dπ₯.
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And so now we have differentiated the function given in the question with respect to π₯ once.
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To obtain that dπ¦ by dπ₯ timesed by cos of π¦ minus two sin of two π₯ is equal to zero.
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And so now weβre ready to differentiate with respect to π₯ for a second time.
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Now we have d by dπ₯ of dπ¦ by dπ₯ cos of π¦ minus two sin of two π₯ is equal to d by dπ₯ of zero.
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And now, again on the right-hand side, we have d by dπ₯ of zero, which is d by dπ₯ of a constant and so will be equal to zero.
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Similarly to the first time, we can split this differentiation up, to give us d by dπ₯ of dπ¦ by dπ₯ times cos of π¦ minus d by dπ₯ of two sin of two π₯ is equal to zero.
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Letβs start by evaluating the left-hand differential.
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In order to evaluate this, weβll need to use the product rule since weβre differentiating a product, which is dπ¦ by dπ₯ multiplied by cos of π¦.
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Now the product rule tells us that if π¦ is equal to some product of variables, so π’ multiplied by π£, then dπ¦ by dπ₯ is equal to π’ multiplied by dπ£ by dπ₯ plus π£ multiplied by dπ’ by dπ₯.
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Therefore, when we differentiate our product of dπ¦ by dπ₯ times cos of π¦, we will get dπ¦ by dπ₯ multiplied by d by dπ₯ of cos of π¦ plus cos of π¦ multiplied by d by dπ₯ of dπ¦ by dπ₯.
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Letβs start by evaluating the d by dπ₯ of cos of π¦.
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So this will be using the chain rule, which tells us that d by dπ₯ of π of π¦ is equal to d by dπ¦ of π of π¦ times dπ¦ by dπ₯.
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So d by dπ₯ of cos of π¦ becomes d by dπ¦ of cos of π¦ timesed by dπ¦ by dπ₯.
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Then d by dπ¦ of cos of π¦ is simply negative sin of π¦.
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And so we obtain negative sin of π¦ times dπ¦ by dπ₯.
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Now we can evaluate the differential d by dπ₯ of dπ¦ by dπ₯.
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Here weβre simply differentiating π¦ with respect to π₯ for a second time.
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And so, therefore, this will be the second differential of π¦ with respect to π₯, or d two π¦ by dπ₯ squared.
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So now we have differentiated all of the terms here.
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We can write this out as dπ¦ by dπ₯ squared times negative sin of π¦ plus d two π¦ by dπ₯ squared times cos of π¦.
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So now we have evaluated d by dπ₯ of dπ¦ by dπ₯ cos of π¦, which is the first differential in our equation.
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Next, we need to evaluate d by dπ₯ of two sin of two π₯.
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Here we will again need to use the chain rule.
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If we let π’ equal two π₯, then we obtain that this is equal to d by dπ’ of two sin of π’ times dπ’ by dπ₯.
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And if we differentiate two sin of π’ with respect to π’, we obtain two cos of π’.
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Then we need to multiply this by the dπ’ by dπ₯.
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And we can substitute back in π’ is equal to two π₯.
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And this gives us two cos of two π₯ times d by dπ₯ of two π₯.
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And to differentiate two π₯ with respect to π₯, we simply multiply by the power, which is one, and then decrease the power of π₯ by one, to give us π₯ to the power of zero, which is simply equal to one.
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So this leaves us with just two, which makes this become two cos of two π₯ multiplied by two, which is the same as four cos of two π₯.
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And so we can substitute this evaluated differential back into our equation.
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And so we obtain that if we differentiate our equation given in the question twice with respect to π₯, we get d two π¦ by dπ₯ squared times cos of π¦ minus dπ¦ by dπ₯ squared timesed by sin of π¦ minus four cos of two π₯ is equal to zero.
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Now in order to reach what weβre trying to prove, we can look at the d two π¦ by dπ₯ squared terms in both the equation that weβre trying to prove and the equation which we just found.
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And we notice that the d two π¦ by dπ₯ squared term in what weβre trying to prove is not being multiplied by anything.
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However, our d two π¦ by dπ₯ squared term is being multiplied by cos of π¦.
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And so, therefore, we should divide the whole equation by cos of π¦.
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And so we obtain that d two π¦ by dπ₯ squared minus dπ¦ by dπ₯ all squared times sin of π¦ over cos of π¦ minus four cos of two π₯ over cos of π¦ is equal to zero.
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We can say that sin of π¦ over cos of π¦ is equal to tan of π¦.
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And we can also say that one over cos of π¦ is equal to sec of π¦.
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So the four cos of two π₯ over cos of π¦ term becomes four cos of two π₯ times sec of π¦.
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And now we can see that our equation is very, very similar to what weβre trying to prove.
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All we need to do is add four cos of two π₯ times sec of π¦ to both sides of the equation, which leaves us with exactly what weβre trying to prove.
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Which is d two π¦ by dπ₯ squared minus dπ¦ by dπ₯ all squared times tan of π¦ is equal to four cos of two π₯ times sec of π¦.
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And therefore, we have completed this question.