### Video Transcript

Find dπ¦ by dπ₯, given that π¦ is equal to the natural logarithm of negative eight π₯ squared divided by seven π₯ minus three.

The question wants us to find dπ¦ by dπ₯. Thatβs the first derivative of π¦ with respect to π₯. And we can see that π¦ is the natural logarithm of a rational function. So, one way of finding dπ¦ by dπ₯ would be to use the quotient rule to differentiate our rational function. We could then use the chain rule to find dπ¦ by dπ₯. And this would work. However, thereβs a simpler method in this case.

Weβre going to use our laws of logarithms to simplify our expression. The first thing weβll do is bring our multiplication by negative one into our denominator. This gives us the natural logarithm of eight π₯ squared divided by three minus seven π₯. Next, we want to use the fact that the log of π minus the log of π is equal to the log of π divided by π. Weβll use this to write our logarithm of a quotient as the difference between two logarithms.

Doing this, we get the natural logarithm of eight π₯ squared minus the natural logarithm of three minus seven π₯. And we can now see both terms in this expression for π¦ are a composition of two functions. So, weβll differentiate this by using the chain rule. And we recall the chain rule tells us, for functions π and π, the derivative of π composed with π is equal to π prime of π₯ times π prime evaluated at π of π₯.

So, letβs start by differentiating our first function. Weβll set π to be the natural logarithm function and π to be eight π₯ squared. So, we now have π of π is equal to the natural logarithm of π. We need to find an expression for π prime of π. And we can differentiate this because we know the derivative of the natural logarithm of π₯ is equal to one divided by π₯. This gives us π prime of π is equal to one over π.

Next, we need to find an expression for π prime of π₯. We know that π of π₯ is eight π₯ squared. So, we can differentiate this by using the power rule for differentiation. We multiply by the exponent of π₯, which is two, and reduce this exponent by one. So, we get π prime of π₯ is equal to 16π₯. So, now, weβre ready to evaluate the derivative of our first term. By the chain rule, this is equal to π prime of π₯ times π prime of π. We showed that π prime of π₯ is 16π₯ and π prime of π is one over π.

And remember, π of π₯ is equal to eight π₯ squared. So, weβll substitute this into our expression for our derivative. And this is equal to 16π₯ divided by eight π₯ squared. And we can simplify this. Weβll cancel the shared factor of π₯ in the numerator and the denominator. And weβll divide by both of our numerator and our denominator through by eight. This gives us two divided by π₯. So, we found the derivative of our first term. Itβs equal to two over π₯. Letβs now clear our working and then find the derivative of our second term.

Weβre gonna want to use the chain rule. So, weβll set π to be the natural logarithm function. And weβll set π to be our inner function three minus seven π₯. Again, we get that π evaluated at π is the natural logarithm of π. And again, we know the derivative of the natural logarithm of π₯ is one over π₯. So, π prime of π is one over π. Next, π of π₯ is our inner function three minus seven π₯. And this is a linear function, so its derivative is just the coefficient of π₯, which in this case is negative seven.

Weβre now ready to evaluate the derivative of our second term by using the chain rule. Itβs equal to π prime of π₯ times π prime evaluated at π. And we showed π prime of π₯ is negative seven and π prime of π is one over π. And again, we know that π of π₯ is equal to three minus seven π₯. So, weβll substitute this into our derivative. And this gives us negative seven divided by three minus π₯. Weβll actually multiply both of our numerator and our denominator through by negative one. And this give us seven divided by seven π₯ minus three.

So, weβve now differentiated both terms in our expression for π¦. This means we can find dπ¦ by dπ₯. So, dπ¦ by dπ₯ will be the difference between our derivatives. Thatβs two over π₯ minus seven divided by seven π₯ minus three. And we could leave our answer like this. However, weβll combine this into one rational function. Cross multiplying and then combining our fractions, we get two times seven π₯ minus three minus seven π₯ all divided by π₯ times seven π₯ minus three.

Multiplying out the parentheses in our numerator, we get 14π₯ minus six minus seven π₯. And 14π₯ minus seven π₯ is equal to seven π₯. So, our numerator simplifies to give us seven π₯ minus six. The last thing weβll do is distribute the π₯ in our denominator over our parentheses. This gives us seven π₯ squared minus three π₯. And this is our final answer. Therefore, weβve shown if π¦ is equal to the natural logarithm of negative eight π₯ squared divided by seven π₯ minus three, then dπ¦ by dπ₯ is equal to seven π₯ minus six all divided by seven π₯ squared minus three π₯.