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

If sin (𝑦) + cos (2π‘₯) = 0, prove that d²𝑦/dπ‘₯Β² βˆ’ (d𝑦/dπ‘₯)Β² tan (𝑦) = 4 cos (2π‘₯) sec (𝑦).

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Video Transcript

If sin of 𝑦 plus cos of two π‘₯ equals zero, prove that d two 𝑦 by dπ‘₯ squared minus d𝑦 by dπ‘₯ squared times tan of 𝑦 is equal to four cos of two π‘₯ times sec of 𝑦.

In order to start answering this question, let’s look at what we’re trying to prove. And we can see that we have a d two 𝑦 by dπ‘₯ squared term and a d𝑦 by dπ‘₯ term. And these tell us that there must be a differential with respect to π‘₯. And that d two 𝑦 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 we’ve just found from what we’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, let’s quickly recall the chain rule. The chain rule tells us that d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 but dπ‘₯, where 𝑒 is another variable. And this makes it easier to differentiate a function within a function.

Now let’s try to differentiate our equation. Now since there are terms which are differentiated with respect to π‘₯ in the thing we’re trying to prove, we’ll differentiate with respect to π‘₯ here. And we’re left with d by dπ‘₯ of sin of 𝑦 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 𝑦 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 𝑒 is equal to two π‘₯, then the chain rule tells us that this differential becomes d by d𝑒 of cos of 𝑒 times d𝑒 by dπ‘₯. Then cos of 𝑒 differentiates to negative sin of 𝑒. And we still need to multiply by the d𝑒 by dπ‘₯.

And now we can substitute back in 𝑒 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 𝑦 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 𝑦 is equal to 𝑓 of 𝑦 into the equation for the chain rule, we end up with d by dπ‘₯ of 𝑓 of 𝑦 is equal to d by d𝑒 of 𝑓 of 𝑦 times d𝑒 by dπ‘₯. However, we’re allowed to choose our variable 𝑒. And so we can let 𝑒 be equal to 𝑦. And so we end up with d by dπ‘₯ of 𝑓 of 𝑦 is equal to d by d𝑦 of 𝑓 of 𝑦 timesed by d𝑦 by dπ‘₯. And so we can apply this to the function we’re trying to resolve. So that’s d by dπ‘₯ of sin of 𝑦.

Now in our case, 𝑓 of 𝑦 is equal to sin of 𝑦. This gives us that d by dπ‘₯ of sin of 𝑦 is equal to d by d𝑦 of sin of 𝑦 times d𝑦 by dπ‘₯. Then d by d𝑦 of sin of 𝑦 is simply cos of 𝑦. And so we get that d by dπ‘₯ of sin of 𝑦 is equal to cos of 𝑦 times d𝑦 by dπ‘₯. And so now we have differentiated the function given in the question with respect to π‘₯ once. To obtain that d𝑦 by dπ‘₯ timesed by cos of 𝑦 minus two sin of two π‘₯ is equal to zero. And so now we’re ready to differentiate with respect to π‘₯ for a second time.

Now we have d by dπ‘₯ of d𝑦 by dπ‘₯ cos of 𝑦 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 d𝑦 by dπ‘₯ times cos of 𝑦 minus d by dπ‘₯ of two sin of two π‘₯ is equal to zero.

Let’s start by evaluating the left-hand differential. In order to evaluate this, we’ll need to use the product rule since we’re differentiating a product, which is d𝑦 by dπ‘₯ multiplied by cos of 𝑦. Now the product rule tells us that if 𝑦 is equal to some product of variables, so 𝑒 multiplied by 𝑣, then d𝑦 by dπ‘₯ is equal to 𝑒 multiplied by d𝑣 by dπ‘₯ plus 𝑣 multiplied by d𝑒 by dπ‘₯. Therefore, when we differentiate our product of d𝑦 by dπ‘₯ times cos of 𝑦, we will get d𝑦 by dπ‘₯ multiplied by d by dπ‘₯ of cos of 𝑦 plus cos of 𝑦 multiplied by d by dπ‘₯ of d𝑦 by dπ‘₯.

Let’s start by evaluating the d by dπ‘₯ of cos of 𝑦. So this will be using the chain rule, which tells us that d by dπ‘₯ of 𝑓 of 𝑦 is equal to d by d𝑦 of 𝑓 of 𝑦 times d𝑦 by dπ‘₯. So d by dπ‘₯ of cos of 𝑦 becomes d by d𝑦 of cos of 𝑦 timesed by d𝑦 by dπ‘₯. Then d by d𝑦 of cos of 𝑦 is simply negative sin of 𝑦. And so we obtain negative sin of 𝑦 times d𝑦 by dπ‘₯.

Now we can evaluate the differential d by dπ‘₯ of d𝑦 by dπ‘₯. Here we’re simply differentiating 𝑦 with respect to π‘₯ for a second time. And so, therefore, this will be the second differential of 𝑦 with respect to π‘₯, or d two 𝑦 by dπ‘₯ squared. So now we have differentiated all of the terms here. We can write this out as d𝑦 by dπ‘₯ squared times negative sin of 𝑦 plus d two 𝑦 by dπ‘₯ squared times cos of 𝑦. So now we have evaluated d by dπ‘₯ of d𝑦 by dπ‘₯ cos of 𝑦, 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 𝑒 equal two π‘₯, then we obtain that this is equal to d by d𝑒 of two sin of 𝑒 times d𝑒 by dπ‘₯. And if we differentiate two sin of 𝑒 with respect to 𝑒, we obtain two cos of 𝑒. Then we need to multiply this by the d𝑒 by dπ‘₯. And we can substitute back in 𝑒 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 𝑦 by dπ‘₯ squared times cos of 𝑦 minus d𝑦 by dπ‘₯ squared timesed by sin of 𝑦 minus four cos of two π‘₯ is equal to zero.

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

And so we obtain that d two 𝑦 by dπ‘₯ squared minus d𝑦 by dπ‘₯ all squared times sin of 𝑦 over cos of 𝑦 minus four cos of two π‘₯ over cos of 𝑦 is equal to zero. We can say that sin of 𝑦 over cos of 𝑦 is equal to tan of 𝑦. And we can also say that one over cos of 𝑦 is equal to sec of 𝑦. So the four cos of two π‘₯ over cos of 𝑦 term becomes four cos of two π‘₯ times sec of 𝑦.

And now we can see that our equation is very, very similar to what we’re trying to prove. All we need to do is add four cos of two π‘₯ times sec of 𝑦 to both sides of the equation, which leaves us with exactly what we’re trying to prove. Which is d two 𝑦 by dπ‘₯ squared minus d𝑦 by dπ‘₯ all squared times tan of 𝑦 is equal to four cos of two π‘₯ times sec of 𝑦. And therefore, we have completed this question.

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