Video: Differentiating a Combination of Logarithmic Functions Using the Chain Rule at a Point

If 𝑓(𝑥) = 3 ln (2𝑥 + 4 ln 𝑥), find 𝑓′(1).

03:47

Video Transcript

If 𝑓 of 𝑥 is equal to three ln multiplied by two 𝑥 plus four ln 𝑥, find the derivative of 𝑓 of 𝑥 when 𝑥 is equal to one.

Now, the first thing we can do if we’re looking to differentiate our function is take our constant term out because this wouldn’t affect the differentiation. So we’ve now got three multiplied by and then the derivative of ln multiplied by two 𝑥 plus four ln 𝑥. Now in order to solve the problem and work out the derivative of the expression, we’re gonna have to use one of our derivative rules. And this is one that’s based around our natural logarithm, which is ln.

So we know that if we want to find the derivative of ln — and in this case, we’re gonna say 𝑢 of 𝑥, so just a function — then it’s equal to one over 𝑢 of 𝑥 multiplied by the derivative of 𝑢 of 𝑥. So therefore, what we’re gonna get is three multiplied by and then we’ve got one over two 𝑥 plus four ln 𝑥. And that’s because that was our 𝑢 of 𝑥. So that was our function when we’re looking at the rule for differentiating something like this. Then, we multiply this by the derivative of two 𝑥 plus four ln 𝑥.

And as we know with differentiation, what we can do is when we’re differentiating an expression like this is we can differentiate each term separately. So we can differentiate two 𝑥 and we can differentiate four ln 𝑥. So first of all, if we differentiate two 𝑥, we’re just gonna get two. And that’s because what we do is we multiply the coefficient by the exponent. So we’d have two multiplied by one which is just two. And then, we reduce exponent from one to zero. So we get two multiplied by one which is just two.

And then, what we’re gonna do to be able to differentiate the second term is use another one of our rules. And that is if we have derivative of ln 𝑥, it’s just equal to one over 𝑥. So if we’re looking to differentiate for ln 𝑥, the first thing we can do — as normal — is take the four, the constant, out. So we have four multiplied by the derivative of ln 𝑥, which is just gonna give us four over 𝑥. And that’s because we’re gonna have four multiplied by one over 𝑥 which is four over 𝑥. So what this does is that it leaves us with an expression. And this expression is three multiplied by two because that was the derivative of two 𝑥 plus four over 𝑥 because that was the derivative of four ln 𝑥. And this is all over two 𝑥 plus four ln 𝑥.

Now, at this stage, we’d think about maybe simplifying. But there’s no need because what we’re trying to find is the value of this expression, so the value of the derivative, when 𝑥 is equal to one. And therefore, to do this, what we need to do is we need to substitute in one for 𝑥 at every point in our derivative expression. So when we do that, we get three multiplied by two plus four over one over two multiplied by one plus four ln one. So then, what we’re gonna get is 18 over two. And that’s because we got three multiplied by two plus four. Well, two plus four is six. Three sixes are 18. Then I’ll need denominator: we’ve got two multiplied by one which is two then add four ln one, well ln one is just equal to zero. So we’re left with two on the denominator. So therefore, this is gonna give us a final answer of nine.

So therefore, we can say that if function 𝑓 of 𝑥 is equal to three ln two 𝑥 plus four ln 𝑥, then the first derivative value when 𝑥 is equal to one is going to be nine.

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