### Video Transcript

Given that π¦ is equal to negative four times π₯ times the cos of two π₯ plus four times the sin of two π₯, find d squared π¦ by dπ₯ squared at π₯ is equal to five π by two.

Weβre given a function of π¦ which is a function of π₯ involving a product and the cos and sin of two π₯. And weβre asked to find the second derivative with respect to π₯ at π₯ is equal to five π divided by two. So weβre going to need to differentiate π¦ with respect to π₯ twice and then evaluate our result at π₯ is equal to five π by two. Since the derivative of the sum is the sum of the derivatives, we can split our function into two. And so the first derivative is d by dπ₯ of negative four π₯ cos two π₯ plus d by dπ₯ of four sin two π₯.

If we look at our first term, weβre differentiating the product negative four π₯ times the cos of two π₯. And using the notation dπ by dπ₯ is equal to π prime, we can use the product rule for differentiation. This says, for a function π of π₯ equal to π’ of π₯ times π£ of π₯, that π prime is equal to π’π£ prime is equal to π’ times π£ prime plus π’ prime times π£.

So if we let π’ equal negative four π₯ and π£ equal cos two π₯, within our first term and to differentiate π’, we can use the power rule for differentiation. This says that for a function of the form π times π₯ raised to the power π where π and π are constants, the derivative with respect to π₯ is π times π times π₯ raised to the power π minus one. That is, we multiply by the exponent of π₯ and subtract one from the exponent. In the case of π’ is equal to negative four π₯, our exponent is one so that the derivative of π’ with respect to π₯ is negative four times one times π₯ raised to the power one minus one, which is negative four times π₯ raised to the power zero. And since anything to the power zero is equal to one, π’ prime is equal to negative four.

Now to differentiate π£ of π₯, which is cos two π₯, weβre going to use the result that the derivative with respect to π₯ of cos ππ₯ is equal to cos ππ₯ prime, which is negative π times the sin of ππ₯. In our case, the constant π is equal to two so that π£ prime is negative two sin two π₯. And now we can use the product rule to differentiate our first term. And thatβs π’, which is negative four π₯, times dπ£ by dπ₯, which is negative two sin two π₯, plus dπ’ by dπ₯, which is negative four, times π£, which is cos two π₯. And distributing our parentheses, this gives us eight π₯ times sin two π₯ minus four cos two π₯ plus the derivative of our second term.

Now, to differentiate our second term, we can use the result that the derivative with respect to π₯ of sin ππ₯ is π cos ππ₯. And again, in our case, the constant π is equal to two. And the derivative of our second term is four times two times the cos of two π₯, that is, eight cos two π₯. And now we can collect our terms negative four cos two π₯ and eight cos two π₯. And our first derivative dπ¦ by dπ₯ is equal to eight π₯ sin two π₯ plus four cos two π₯.

So we found the first derivative of π¦ with respect to π₯, but in fact, weβre asked to find the second. And that means we have to differentiate the first derivative. We can actually use exactly the same methods and differentiate term by term. And to differentiate our first term, which is eight π₯ times sin two π₯, a product, we can let π’ of π₯ equal to eight π₯ and π£ of π₯ equals sin two π₯.

π’ prime is equal to eight by the power rule, and π£ prime is equal to two cos two π₯ by our result here. So now, using the product rule, the derivative of our first term is eight π₯, which is π’, times two cos two π₯, which is π£ prime, plus eight, which is π’ prime, times sin two π₯, which is π£. And we add on the derivative of our second term, four cos two π₯, to get d squared π¦ by dπ₯ squared. If we then multiply out our second derivative, itβs 16π₯ cos two π₯ plus eight sin two π₯ plus the derivative of four cos two π₯. And to differentiate our second term, we can use the known result that d by dπ₯ of cos ππ₯ is negative π sin ππ₯, where again, in our case, the constant π is equal to two. And so the derivative of our second term is negative four times two sin two π₯, which is negative eight sin two π₯.

Now in our second derivative, we have a positive and a negative eight sin two π₯, so these can cancel, which leaves us with d squared π¦ by dπ₯ squared is equal to 16π₯ times the cos of two π₯. Remember, though, that weβre asked to find d squared π¦ by dπ₯ squared at π₯ is equal to five π divided by two. So now if we make some space, so now making some space, we want to substitute π₯ is equal to five π by two into our d squared π¦ by dπ₯ squared. This gives us 16 times five π by two times the cos of two times five π by two. And we can cancel our twos, giving us eight times five π times the cos of five π. That is 40π times the cos of five π.

It remains then to evaluate the cos of five π. And to do this, we can use the result that for an integer π, the cos of ππ is negative one raised to the πth power. So that if π is an even integer, the cos of ππ is positive one, and if π is an odd integer, the cos of ππ is negative one. In the case of our second derivative, we have π equal to five, which is an odd number, so the cos of five π is negative one. And so d squared π¦ by dπ₯ squared at π₯ is equal to five π by two is negative 40π. And so for π¦ is equal to negative four π₯ times cos two π₯ plus four times sin two π₯, d squared π¦ by dπ₯ squared at π₯ is equal to five π by two is negative 40π.