Question Video: Deriving the Inverse Function of an Exponential Function | Nagwa Question Video: Deriving the Inverse Function of an Exponential Function | Nagwa

Question Video: Deriving the Inverse Function of an Exponential Function Mathematics

Find d𝑦/dπ‘₯, given that 𝑦 = ln (2π‘₯ + 7)Β³.

02:14

Video Transcript

Find d𝑦 by dπ‘₯, given that 𝑦 is equal to the natural log of two π‘₯ plus seven cubed.

In this question, we have a function of a function, in other words, a composite function. We can use the chain rule to find the derivative of a composite function, though since we’re working with the natural logarithm, it’s sensible to first consider whether there is anything we can do to manipulate our expression before differentiating. We recall the general result for the natural logarithm of a power. The natural logarithm of π‘₯ to the 𝑦th power is 𝑦 times the natural logarithm of π‘₯. And we see that we can now rewrite our equation as 𝑦 equals three times the natural logarithm of two π‘₯ plus seven. This is great because we know that the derivative of a constant multiple of an expression in π‘₯ is equal to the multiple of the derivative of that expression. In other words, the derivative of three times the natural log of two π‘₯ plus seven is equal to three times the derivative of the natural log of two π‘₯ plus seven.

But how do we differentiate the natural log of two π‘₯ plus seven? We’re going to quote the general result for the derivative of the natural logarithm and we’re going to use the chain rule. The chain rule says that if 𝑦 is a function in 𝑒 and 𝑒 itself is a function in π‘₯, then the derivative of 𝑦 with respect to π‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. And we also know that the derivative of the natural log of π‘₯ is simply one over π‘₯. So we let 𝑒 be equal to two π‘₯ plus seven such that 𝑦 is equal to the natural log of 𝑒. We know that d𝑒 by dπ‘₯ is two and d𝑦 by d𝑒 is one over 𝑒. The derivative then of the natural log of two π‘₯ plus seven is the product of these. It’s two times one over 𝑒. But we replace 𝑒 with two π‘₯ plus seven.

And we see that the derivative of the natural log of two π‘₯ plus seven is two over two π‘₯ plus seven. We recall we said that the derivative of three times the natural log of two π‘₯ plus seven is three times the derivative of the natural log of two π‘₯ plus seven. So the derivative is three times two over two π‘₯ plus seven, which is simply six over two π‘₯ plus seven.

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