Question Video: Differentiating a Composition of Exponential and Trigonometric Functions Using Logarithmic Differentiation Mathematics

Video Transcript

Given that 𝑦 is equal to two raised to the power negative nine 𝑒 to the nine 𝑥 plus sin 𝑥, determine d𝑦 by d𝑥.

We’re given quite a complicated function 𝑦, where the exponent is a function of 𝑥. It is possible to use the chain rule to differentiate 𝑦, but we’ll use an alternative method. That is logarithmic differentiation. And how does this work? Well, if we have a function 𝑦 which is a function of 𝑥, our first step is to apply the natural logarithm to both sides so that the natural logarithm of 𝑦 is the natural logarithm of 𝑓 of 𝑥, remembering that the natural logarithm means the logarithm to the base 𝑒, where 𝑒 is Euler’s number which is approximately 2.71828 and so on.

In our case, our function 𝑦 is two raised to the power negative nine 𝑒 to the nine 𝑥 plus sin 𝑥. And in order for this first step to be valid, that is, taking natural logarithms, we need to specify that 𝑦 is greater than zero. That’s because the logarithm function doesn’t exist for negative values and the logarithm of zero is undefined. And since two raised to any power is positive, our 𝑦 is indeed greater than zero.

We’ve taken the natural logarithm, but our function looks more complicated than it did to begin with. But this is where the laws of logarithms come in useful. Our second step is to use the laws of logarithms to simplify or expand our right-hand side. And since our original function has a complicated exponent, we can use the power rule for logarithms to simplify this. The power rule says that log to the base 𝑎 of 𝑏 raised to the power 𝑐 is 𝑐 times log to the base 𝑎 of 𝑏. That is, we bring our exponent in front of the logarithm and multiply.

In our problem, our exponent is negative nine 𝑒 to the nine 𝑥 plus sin 𝑥. And applying the power rule, we bring this to the front and multiply by it. And so we have the natural logarithm of 𝑦 is equal to negative nine 𝑒 to the nine 𝑥 plus sin 𝑥 times the natural logarithm of two. So on our right-hand side, we have the constant, the natural logarithm of two, multiplied by negative nine 𝑒 to the nine 𝑥 plus sin 𝑥 which is a function of 𝑥. And we know how to differentiate this.

And this brings us to our third step in logarithmic differentiation, that is, differentiate both sides with respect to 𝑥. Making some room, on our right-hand side, we can take the natural logarithm of two outside since it’s a constant. And since the derivative of a sum is the sum of the derivatives, we have the natural logarithm of two times d by d𝑥 of negative nine 𝑒 to the nine 𝑥 plus d by d𝑥 of sin 𝑥. And again, since negative nine this time is a constant, we can take this out front. Now, making some room, on our right-hand side, the first thing we want to differentiate is 𝑒 to the power nine 𝑥. And for this, we can use the known result that for 𝑢, a differentiable function of 𝑥, d by d𝑥 of 𝑒 to the power 𝑢 is equal to d𝑢 by d𝑥 times 𝑒 to the power 𝑢.

In our case, we have 𝑢 is equal to nine 𝑥 and d𝑢 by d𝑥 is equal to nine. So we have d by d𝑥 of 𝑒 to the power nine 𝑥 is nine, which is d𝑢 by d𝑥, times 𝑒 to the power nine 𝑥. Our second term on the right-hand side is d by d𝑥 of sin 𝑥. And we know that d by d𝑥 of sin 𝑥 is cos 𝑥. On our left-hand side, we have the derivative with respect to 𝑥 of the natural logarithm of 𝑦. But remember that 𝑦 is a function of 𝑥. And again, we can use the known result to differentiate this, since the derivative with respect to 𝑥 of the natural logarithm of 𝑢 where 𝑢 is a differentiable function of 𝑥 is one over 𝑢 times d𝑢 by d𝑥 for 𝑢 greater than zero.

On our left-hand side then, we have one over 𝑦 times d𝑦 by d𝑥. And that’s equal on our right-hand side to the natural algorithm of two times negative nine times nine 𝑒 to the nine 𝑥 plus cos 𝑥. Negative nine times nine is negative 81, and we’re almost but not quite finished.

Our final step is to solve for d𝑦 by d𝑥. We can do this by noting that we have a factor of one over 𝑦 on our left-hand side. And to eliminate this, we can multiply both sides by 𝑦. On our left-hand side, these cancel to one, and on our right-hand side, we’re going to have to reintroduce our original function 𝑦 so that if 𝑦 is equal to two to the power negative nine 𝑒 to the nine 𝑥 plus sin 𝑥, d𝑦 by d𝑥 is two raised to the power negative nine 𝑒 to the nine 𝑥 plus sin 𝑥 times negative 81𝑒 to the power nine 𝑥 plus cos 𝑥 all multiplied by the natural logarithm of two.