# Question Video: Differentiating a Combination of Logarithm Arithmic and Polynomial Functions Using the Quotient Rule Mathematics • Higher Education

Find d𝑦/d𝑥, given that 𝑦 = 9𝑥/ln 9𝑥.

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### Video Transcript

Find d𝑦 by d𝑥 given that 𝑦 is equal to nine 𝑥 divided by the natural logarithm of nine 𝑥.

The question wants us to find d𝑦 by d𝑥. That’s the first derivative of 𝑦 with respect to 𝑥. And we can see that 𝑦 is the quotient of two functions. It’s the quotient of nine 𝑥 and the natural logarithm of nine 𝑥. So we’ll find this derivative by using the quotient rule. We recall the quotient rule tells us if 𝑦 is the quotient of two functions 𝑢 over 𝑣, then d𝑦 by d𝑥 is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all divided by 𝑣 squared.

So to use the quotient rule, we’ll start by setting 𝑢 of 𝑥 to be the function in our numerator, that’s nine 𝑥, and 𝑣 of 𝑥 to be the function in our denominator, that’s the natural logarithm of nine 𝑥. And to apply the quotient rule, we’re going to need to find expressions for d𝑢 by d𝑥 and d𝑣 by d𝑥. Let’s start with finding d𝑢 by d𝑥. That’s the derivative of nine 𝑥 with respect to 𝑥. And nine 𝑥 is just a linear function. So its derivative is the coefficient of 𝑥, which, in this case, is nine.

Let’s now find an expression for d𝑣 by d𝑥. That’s the derivative of the natural logarithm of nine 𝑥 with respect to 𝑥. And we can do this by using one of our standard derivative results for logarithmic functions. For any positive constant 𝑎, the derivative of the natural logarithm of 𝑎𝑥 with respect to 𝑥 is equal to one divided by 𝑥. So in our case, the derivative of the natural logarithm of nine 𝑥 with respect to 𝑥 is equal to one divided by 𝑥. So we’re now ready to find the d𝑦 by d𝑥 by using the quotient rule.

The quotient rule tells us d𝑦 by d𝑥 will be equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 divided by 𝑣 squared. Substituting in our expressions for 𝑢, 𝑣, d𝑢 by d𝑥, and d𝑣 by d𝑥, we get d𝑦 by d𝑥 is equal to the natural logarithm of nine 𝑥 multiplied by nine minus nine 𝑥 times one over 𝑥 all divided by the natural logarithm of nine 𝑥 squared. And we can simplify this expression. First, we’ll cancel 𝑥 multiplied by one over 𝑥.

Next, we want to take out a factor of nine in our numerator. And this gives us nine times the natural logarithm of nine 𝑥 minus one all divided by the natural logarithm of nine 𝑥 squared. And this is our final answer. Therefore, we’ve shown if 𝑦 is equal to nine 𝑥 divided by the natural logarithm of nine 𝑥, then d𝑦 by d𝑥 is equal to nine times the natural logarithm of nine 𝑥 minus one all divided by the natural logarithm of nine 𝑥 squared.