Video: Differentiating a Composition of Root and Rational Functions Using Logarithmic Differentiation

Using logarithmic differentiation, determine the derivative of 𝑦 = √((π‘₯ + 1)/(2π‘₯⁴ βˆ’ 2)).

04:44

Video Transcript

Using logarithmic differentiation, determine the derivative of 𝑦 is equal to the square root of π‘₯ plus one over two π‘₯ to the power four minus two.

We’re given what looks to be a fairly complicated function 𝑦, which is the square root of a quotient function. And we can write this as π‘₯ plus one over two π‘₯ to the of power four minus two all raised to the power of a half since an exponent of a half is the same thing as a square root. Now we could treat this as a function of a function and use the chain rule and the quotient rule to differentiate. However, we’re specifically asked to use logarithmic differentiation.

But how does this work? Well, let’s see. Suppose we have a function 𝑦 is equal to 𝑓 of π‘₯. Then the first thing we do is apply the natural logarithm to both sides. We then have the natural logarithm of 𝑦 is equal to the natural logarithm of 𝑓 of π‘₯, recalling that the natural logarithm means the log to the base 𝑒, where 𝑒 is Euler’s number, which is approximately 2.71828 and so on. And for this to be valid, we need to specify that 𝑦 is greater than zero, since the log of zero is undefined and the log doesn’t exist for negative values.

If we did want to include negative values, we use the absolute values of 𝑦 and 𝑓 of π‘₯ and then specify that 𝑦 is not equal to zero. Our second step in logarithmic differentiation is to use the laws of logarithms to expand the natural log of 𝑓 of π‘₯. This should give us something that we can easily differentiate. And so we differentiate on both sides with respect to π‘₯. And once we’ve done this, we can solve for 𝑦 prime. That’s d𝑦 by dπ‘₯. So let’s begin working through our steps with our function 𝑦 by taking the natural logarithms on both sides.

And in our case, we can specify that 𝑦 is the positive square root. Next, we use the laws of logarithms to expand our right-hand side. And the first law we can use since we have an exponent of a half on our right-hand side is the power rule for logarithms. This says that log to the base π‘Ž of 𝑏 raised to the power 𝑐 is 𝑐 times log to the base π‘Ž of 𝑏. That is, we bring our exponent to the front and multiply by it. In our case then, this means we bring our half down to the front.

Now the argument of our natural logarithm on the right-hand side is a quotient. And we can simplify this by using the quotient rule for logarithms that says log to the base π‘Ž of 𝑏 over 𝑐 is the log to the base π‘Ž of 𝑏 minus log to the base π‘Ž of 𝑐. In our case, π‘₯ plus one corresponds to 𝑏 and two π‘₯ to the power of four minus two corresponds to 𝑐. And so we have one over two times the natural logarithm of π‘₯ plus one minus the natural logarithm of two π‘₯ to the power of four minus two. So now we have something manageable on our right-hand side that we know how to differentiate quite easily.

Now we want the derivative with respect to π‘₯. And on our left-hand side, we’re differentiating a function of a function because, of course, 𝑦 is a function of π‘₯. And on our right-hand side, the derivative of a sum is the sum of the derivatives. So we can split this into two derivatives and take the constant a half out front. And on both sides, we’re going to use the known result that d by dπ‘₯ of the natural logarithm of 𝑒, where 𝑒 is a differentiable function of π‘₯, is one over 𝑒 times d𝑒 by dπ‘₯. And that’s for 𝑒 greater than zero.

On our left-hand side, this gives us one over 𝑦 d𝑦 by dπ‘₯. And on our right-hand side, in our first term, if we have 𝑒 is equal to π‘₯ plus one and d𝑒 by dπ‘₯ is equal to one, then d by dπ‘₯ of the natural logarithm of π‘₯ plus one is one times one over π‘₯ plus one. Similarly, in our second term, if we let 𝑒 equal to two π‘₯ to the power of four minus two, then d𝑒 by dπ‘₯ is eight times π‘₯ to the power three. And so the derivative with respect to π‘₯ of the natural logarithm of two π‘₯ raised to power of four minus two is eight π‘₯ cubed times one over two π‘₯ to the power of four minus two.

And writing this out a little more neatly, we have one over 𝑦 d𝑦 by dπ‘₯ on our left-hand side is equal to one over two times one over π‘₯ plus one minus eight π‘₯ cubed over two π‘₯ to the power of four minus two. In the second term, on our right-hand side, in the denominator, we have a common factor of two, which we can cancel with the factor two in the numerator. And we can also bring our half inside our parentheses. And we can note again that in the second term in our right-hand side, we have a common factor of two in the numerator and denominator. And so we have one over 𝑦 d𝑦 by dπ‘₯ is one over two π‘₯ plus two minus two π‘₯ to the power of three over π‘₯ to the power of four minus one.

Now making some room, we can note that we have a factor of one over 𝑦 on the left-hand side, so we’re not quite finished. To eliminate this, we can multiply both sides by 𝑦. And the 𝑦’s on the left-hand side will then cancel. And if we then substitute back in our function 𝑦, we have d𝑦 by dπ‘₯ is as shown. And if we rewrite this as a square root as it’s given in the question, we have d𝑦 by dπ‘₯, so that’s the derivative of 𝑦 with respect to π‘₯, is the square root of π‘₯ plus one over two π‘₯ to the power of four minus two times one over two π‘₯ plus two minus two π‘₯ cubed over π‘₯ to the power of four minus one.

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