### Video Transcript

If π¦ is equal to cos to the seventh power of π₯ divided by nine π₯ plus five, find the first derivative of π¦ with respect to π₯.

In this question, weβre asked to find the first derivative of π¦ with respect to π₯. And to do this, letβs start by looking at the function weβre asked to differentiate. First, we can see that our argument is a rational function. Itβs the quotient of two polynomials. So we can differentiate the arguments of our trigonometric function by using the quotient rule. Second, we can see that our outer function is actually the composition of two functions. Weβre raising a trigonometric function to the seventh power. We can differentiate the composition of functions by using the chain rule.

We recall the chain rule tells us the derivative of the composition of two functions π composed of π is equal to π prime of π₯ multiplied by π prime evaluated at π of π₯. And this rule is valid provided π is differentiable at π₯ and π is differentiable at π of π₯.

To apply the chain rule, weβre first going to need to write π¦ as the composition of two functions. And to do this, we need to notice the outermost function is the exponent of the seventh power. So our inner function will be the trigonometric function. π of π₯ will be the cos of π₯ divided by nine π₯ plus five. And then π will be our outer exponential function. π evaluated at π of π₯ is π to the seventh power, or π evaluated at our variable π is π to the seventh power. Therefore, weβve rewritten π¦ to be equal to π evaluated at π of π₯, which therefore means by the chain rule dπ¦ by dπ₯ is equal to π prime of π₯ multiplied by π prime evaluated at π of π₯.

So to apply the chain rule, we need to find expressions for π prime and π prime. Letβs start by finding an expression for π prime. We need to differentiate π to the seventh power with respect to π. We can do this by using the power rule for differentiation. We need to multiply by the exponent of π and then reduce this exponent by one. π prime of π is seven π to the sixth power. We then need to find an expression for π prime of π₯. However, we can see that π of π₯ is the composition of two functions. Itβs the cosine of a rational function. So weβre going to need to apply the chain rule to differentiate π of π₯. Weβll start by calling our inner rational function π’ of π₯. And we can call our outer trigonometric function π£. Then, π of π₯ will be equal to π£ evaluated at π’ of π₯. π£ is a function in π’, and π’ in turn is a function in π₯. So we can differentiate this by using the chain rule.

Once again, to apply the chain rule, weβre going to need to find the derivatives of both of these functions. Letβs start by finding π£ prime of π’. π£ prime of π’ is the derivative of the cos of π’ with respect to π’. And we know the derivative of the cosine function is negative the sine function. π£ prime of π’ is negative sin of π’.

We would then need to differentiate π’ of π₯. However, π’ of π₯ is a rational function. Itβs the quotient of two polynomials. So weβll do this by using the quotient rule. And we recall the quotient rule tells us the derivative of π of π₯ divided by π of π₯ is equal to π prime of π₯ times π of π₯ minus π prime of π₯ times π of π₯ all divided by π of π₯ all squared. And this will be valid for all values of π₯ provided π is differentiable at π₯, π is differentiable at π₯, and π of π₯ is not equal to zero.

To apply this to the rational function π’ of π₯, we need to set π of π₯ to be the function in our numerator and π of π₯ to be the function in our denominator. π of π₯ is equal to π₯, and π of π₯ is equal to nine π₯ plus five. We then need to find the derivative of both of these functions. Since both of these are linear functions, we could do this by using the power rule for differentiation. However, we know the slope of any linear function is the coefficient of π₯. So π prime of π₯ is one and π prime of π₯ is nine.

Now that we found expressions for π prime of π₯ and π prime of π₯, we can substitute our expressions for π prime, π prime, π, and π into the quotient rule to find π’ prime of π₯. Substituting these expressions in, we get π’ prime of π₯ is equal to one multiplied by nine π₯ plus five minus nine π₯ all divided by nine π₯ plus five all squared. And if we distribute and simplify this expression, we see that π’ prime of π₯ is equal to five divided by nine π₯ plus five all squared.

Letβs now clear some space. Weβve now found expressions for π£ prime and π’ prime. We can substitute these into the chain rule to find an expression for π prime of π₯. This is because the chain rule tells us that π prime of π₯ is equal to π’ prime of π₯ multiplied by π£ prime evaluated at π’ of π₯. Substituting in our expressions for π’ prime, π£ prime, and π£, we get that π prime of π₯ is equal to five divided by nine π₯ plus five all squared multiplied by negative sin of π₯ divided by nine π₯ plus five. We can then simplify this expression by writing the entire function divided by nine π₯ plus five all squared. This gives us negative five sin of π₯ divided by nine π₯ plus five all divided by nine π₯ plus five all squared.

Letβs once again clear some space. Now that we found expressions for π prime, π prime, and π of π₯, we can use the chain rule to differentiate π¦. Substituting in our expressions for π prime, π, and π, we get dπ¦ by dπ₯ is equal to negative five times sin of π₯ divided by nine π₯ plus five all divided by nine π₯ plus five all squared all multiplied by seven times cos of π₯ divided by nine π₯ plus five all raised to the sixth power.

And we can simplify this expression. Weβll start by writing the entire function divided by nine π₯ plus five all squared. Next, weβll start simplifying the numerator. We have negative five multiplied by seven, which is negative 35. We then need to multiply this by the sin of π₯ divided by nine π₯ plus five. And then, finally, we can simplify the cos of π₯ divided by nine π₯ plus five all raised to the sixth power by writing the six above the cosine, giving us the following expression, which is our final answer.

Therefore, we were able to show if π¦ is cos to the seventh power of π₯ divided by nine π₯ plus five, then the first derivative of π¦ with respect to π₯ is equal to negative 35 sin of π₯ divided by nine π₯ plus five multiplied by cos to the sixth power of π₯ divided by nine π₯ plus five all divided by nine π₯ plus five all squared.