### Video Transcript

Find the derivative of the function π¦ is equal to the natural logarithm of π₯ raised to the fourth power times the tangent to the fourth power of π₯.

The question wants us to find the derivative of our function π¦. And thereβs actually several different ways which we can approach this. For example, we could think of π¦ as the product of two functions. Itβs the product of the natural logarithm of π₯ raised to the fourth power multiplied by the tangent to the fourth power of π₯. We could then differentiate this by using the product rule. And we could differentiate each of these terms separately by using the chain rule.

And this would work. However, thereβs a second method which involves using our laws of exponents. We can see that both of these factors are raised to the fourth power. And we know, by our laws of exponents, π to the fourth power times π to the fourth power is the same as π times π all raised to the fourth power. So, by using our laws of exponents, we can rewrite π¦ as the natural logarithm of π₯ times the tangent of π₯ all raised to the fourth power.

By setting π of π₯ to be our inner function the natural logarithm of π₯ times the tangent of π₯, we get that π¦ is equal to π of π₯ to the fourth power. In other words, π¦ is the composition of two functions. So, we can find the derivative of π¦ by using the chain rule. Weβll recall the chain rule tells us if π¦ is the composition of two functions π composed with π of π₯, then the derivative of π¦ is equal to π prime of π₯ times the derivative of π evaluated at π of π₯.

In our case, weβre raising π to the fourth power. So, our function π of π is equal to π to the fourth power. And we can differentiate this by using the power rule for differentiation. We multiply by the exponent of π and reduce this exponent by one. This gives us four π cubed. Therefore, by the chain rule, we have that π¦ prime is equal to π prime of π₯ times four π cubed. And remember, we know that π of π₯ is equal to the natural logarithm of π₯ times the tangent of π₯.

So, substituting this in, weβve shown that π¦ prime is equal to π prime of π₯ times four times the natural logarithm of π₯ times the tangent of π₯ cubed. We now need to find an expression for π prime of π₯. And remember, π of π₯ is equal to the natural logarithm of π₯ times the tangent of π₯. Itβs the product of two functions. So, weβll differentiate this by using the product rule. We recall the product rule tells us if we have a function π of π₯ which is the product of two functions π’ of π₯ and π£ of π₯. Then π prime of π₯ is equal to π£ of π₯ times π’ prime of π₯ plus π’ of π₯ times π£ prime of π₯.

So, to differentiate π of π₯, weβll set π’ of π₯ to be the natural logarithm of π₯ and π£ of π₯ to be the tangent of π₯. Now, we need to find expressions for π’ prime of π₯ and π£ prime of π₯. Luckily, both of these are standard derivatives. We know the derivative of the natural logarithm of π₯ with respect to π₯ is one over π₯. And we also know the derivative of the tangent of π₯ with respect to π₯ is equal to the sec squared of π₯.

So, weβre now ready to find an expression for π prime of π₯. By the product rule, this is equal to π£ of π₯ times π’ prime of π₯ plus π’ of π₯ times π£ prime of π₯. Substituting in our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯, we have that π prime of π₯ is equal to the tangent of π₯ times one over π₯ plus the natural logarithm of π₯ times the sec squared of π₯. And weβll simplify this slightly by multiplying the numerator and the denominator of our second term by π₯.

Since our first term also has a denominator of π₯, we can now add these two terms together, giving us that π prime of π₯ is equal to the tangent of π₯ plus π₯ times the natural logarithm of π₯ times the sec squared of π₯ all divided by π₯. We can now substitute this into our expression for π¦ prime of π₯. However, before we do this, weβll move our coefficient of four to the beginning of our expression and our factor of π prime of π₯ to the end of our expression.

Weβre now ready to substitute in our expression for π prime of π₯. So, substituting in our expression for π prime of π₯. Weβve shown that π¦ prime of π₯ is equal to four times the natural logarithm of π₯ times the tangent of π₯ cubed multiplied by the tangent of π₯ plus π₯ times the natural logarithm of π₯ times the sec squared of π₯ divided by π₯.