Question Video: Differentiating a Composition of Logarithmic and Rational Functions Using the Chain Rule | Nagwa Question Video: Differentiating a Composition of Logarithmic and Rational Functions Using the Chain Rule | Nagwa

# Question Video: Differentiating a Composition of Logarithmic and Rational Functions Using the Chain Rule Mathematics • Third Year of Secondary School

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Find dπ¦/dπ₯, given that π¦ = ln(β8π₯Β²/7π₯ β 3).

04:57

### 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₯. Thatβs 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, thereβs a simpler method in this case.

Weβre going to use our laws of logarithms to simplify our expression. The first thing weβll 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. Weβll 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, weβll 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, letβs start by differentiating our first function. Weβll 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, weβre 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, weβll substitute this into our expression for our derivative. And this is equal to 16π₯ divided by eight π₯ squared. And we can simplify this. Weβll cancel the shared factor of π₯ in the numerator and the denominator. And weβll 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. Itβs equal to two over π₯. Letβs now clear our working and then find the derivative of our second term.

Weβre gonna want to use the chain rule. So, weβll set π to be the natural logarithm function. And weβll 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.

Weβre now ready to evaluate the derivative of our second term by using the chain rule. Itβs 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, weβll substitute this into our derivative. And this gives us negative seven divided by three minus π₯. Weβll 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, weβve 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. Thatβs two over π₯ minus seven divided by seven π₯ minus three. And we could leave our answer like this. However, weβll 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 weβll 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, weβve 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₯.

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