# Video: MATH-DIFF-INT-2018-S1-Q04

If sin (π¦) + cos (2π₯) = 0, prove that dΒ²π¦/dπ₯Β² β (dπ¦/dπ₯)Β² tan (π¦) = 4 cos (2π₯) sec (π¦).

09:32

### Video Transcript

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¦.

In order to start answering this question, letβs look at what weβre trying to prove. And we can see that we have a d two π¦ by dπ₯ squared term and a dπ¦ by dπ₯ term. And these tell us that there must be a differential with respect to π₯. 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.

Using this information that weβve just found from what weβre trying to prove, the next logical step will be to differentiate the equation given in the question with respect to π₯ twice. In order to do this, letβs quickly recall the chain rule. The chain rule tells us that dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ but dπ₯, where π’ is another variable. And this makes it easier to differentiate a function within a function.

Now letβs try to differentiate our equation. Now since there are terms which are differentiated with respect to π₯ in the thing weβre trying to prove, weβll differentiate with respect to π₯ here. And weβre left with d by dπ₯ of sin of π¦ plus cos of two π₯ is equal to d by dπ₯ of zero. Now this type of differentiation where we just differentiate a whole equation is called implicit differentiation.

To start off here, we can evaluate the right-hand side, since if you differentiate a constant by anything it will give you zero. So d by π₯ of zero is simply zero. 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.

And now in order to resolve the d by dπ₯ of cos of two π₯, we can use the chain rule. 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₯. Then cos of π’ differentiates to negative sin of π’. And we still need to multiply by the dπ’ by dπ₯.

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₯. 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. And so it becomes two times π₯ to the power of zero, which is simply two. And so now we found the differential of cos of two π₯ with respect to π₯. And it is negative two sin of two π₯.

And now in order to differentiate sin of π¦ with respect to π₯, we will again be using the chain rule. However, we can adapt the chain rule slightly to make this easier for ourselves. 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₯. However, weβre allowed to choose our variable π’. And so we can let π’ be equal to π¦. 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₯. And so we can apply this to the function weβre trying to resolve. So thatβs d by dπ₯ of sin of π¦.

Now in our case, π of π¦ is equal to sin of π¦. 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₯. Then d by dπ¦ of sin of π¦ is simply cos of π¦. And so we get that d by dπ₯ of sin of π¦ is equal to cos of π¦ times dπ¦ by dπ₯. And so now we have differentiated the function given in the question with respect to π₯ once. To obtain that dπ¦ by dπ₯ timesed by cos of π¦ minus two sin of two π₯ is equal to zero. And so now weβre ready to differentiate with respect to π₯ for a second time.

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. 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. 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.

Letβs start by evaluating the left-hand differential. In order to evaluate this, weβll need to use the product rule since weβre differentiating a product, which is dπ¦ by dπ₯ multiplied by cos of π¦. 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₯. 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₯.

Letβs start by evaluating the d by dπ₯ of cos of π¦. 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₯. So d by dπ₯ of cos of π¦ becomes d by dπ¦ of cos of π¦ timesed by dπ¦ by dπ₯. Then d by dπ¦ of cos of π¦ is simply negative sin of π¦. And so we obtain negative sin of π¦ times dπ¦ by dπ₯.

Now we can evaluate the differential d by dπ₯ of dπ¦ by dπ₯. Here weβre simply differentiating π¦ with respect to π₯ for a second time. And so, therefore, this will be the second differential of π¦ with respect to π₯, or d two π¦ by dπ₯ squared. So now we have differentiated all of the terms here. 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¦. So now we have evaluated d by dπ₯ of dπ¦ by dπ₯ cos of π¦, which is the first differential in our equation.

Next, we need to evaluate d by dπ₯ of two sin of two π₯. Here we will again need to use the chain rule. 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₯. And if we differentiate two sin of π’ with respect to π’, we obtain two cos of π’. Then we need to multiply this by the dπ’ by dπ₯. And we can substitute back in π’ is equal to two π₯. And this gives us two cos of two π₯ times d by dπ₯ of two π₯. 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. 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₯.

And so we can substitute this evaluated differential back into our equation. 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.

Now in order to reach what weβre trying to prove, we can look at the d two π¦ by dπ₯ squared terms in both the equation that weβre trying to prove and the equation which we just found. And we notice that the d two π¦ by dπ₯ squared term in what weβre trying to prove is not being multiplied by anything. However, our d two π¦ by dπ₯ squared term is being multiplied by cos of π¦. And so, therefore, we should divide the whole equation by cos of π¦.

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. We can say that sin of π¦ over cos of π¦ is equal to tan of π¦. And we can also say that one over cos of π¦ is equal to sec of π¦. So the four cos of two π₯ over cos of π¦ term becomes four cos of two π₯ times sec of π¦.

And now we can see that our equation is very, very similar to what weβre trying to prove. 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 weβre trying to prove. 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¦. And therefore, we have completed this question.