How do I evaluate the following derivative?

Question

How can I evaluate the derivative $$\left. \frac{d}{dx} \left( x \, \ln(x^5) \right)\right|_{x=1} \, ?$$

I have this math problem, and the biggest thing I do not understand, is which derivative rule do I need to apply, and what does the vertical bar mean in this problem?

Solution $$\left. \frac{d}{dx} \left( x \, \ln(x^5) \right)\right|_{x=1} = 5.$$

Answer 1

$$\frac{d}{dx}(x\ln(x^5))=$$

$$\ln(x^5)\frac{d}{dx}(x)+x\frac{d}{dx}(5\ln(x))=$$

$$\ln(x^5)+x\frac{5}{x}=$$

$$\ln(x^5)+5$$

for $x=1$, it gives $0+5=5$.

Answer 2

As an alternative since $x\ln(x^5)=5x\ln(x)$ by product rule

$$\frac{d}{dx}(x\ln(x^5))=5\frac{d}{dx}(x\ln x)=5\left(1\cdot \ln x+x\cdot \frac1x\right)=5\ln x+5$$

Answer 3

The 'Vertical Bar' simply gives you the point at which you have to evaluate the derivative function.

For Example, let function at hand be $f(x)$ and you are to find its derivative and evaluate it at the point $x=a$

Here's what you would do:

$$\frac{d}{dx}(f(x))|_{x=a}=f'(x)|_{x=a}$$

which just essentially is the value of the derivative function at $x=a$ i.e. $f'(a)$.

Coming to the second part of your question, you're basically using the Product Rule which for two differentiable functions $u=f(x)$ and $v=g(x)$ says that the derivative of the product of two differentiable functions is equal to the sum of the derivative of one function times the other function and the derivative of the other function times the first which in mathematical terms is really stated as follows:

$$\frac{d}{dx}(uv)=u\frac{d}{dx}v+v\frac{d}{dx}u$$

You might also use the Chain Rule if you are not familiar with logarithm properties. The Chain Rule helps us calculate the derivative of a function nested inside another, one like, $f(g(x))$ which in your questions is $ln(x^5)$. Chain Rule mathematically states the following:

$$\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)$$

where $f'(g(x))$ is the derivative of $f(x)$ with the function $g(x)$ as its input and $g'(x)$ is the derivative of $g(x)$, both are w.r.t. x.

However, if you use one of the logarithm properties, you wouldn't need to use the chain rule explicitly. Using this property simplifies things in this case:

$$\ln(x^a)=a\ln(x)$$

Below I'll solve the differentiation problem stated with two different approaches, first using the Chain Rule, then using Logarithm Properties. Notice, however, the Product Rule is being used in both the cases

Note: I'll put the pipe ("vertical bar"), once I'm done differentiating for the sake of convenience.

Approach 1 (Using Chain Rule)

$\frac{d}{dx}{x\ln(x^5)}$

$\Rightarrow x\frac{d}{dx}(\ln(x^5))+\ln(x^5)\frac{d}{dx}(x)$

$\Rightarrow x(\frac{\frac{d}{dx}(x^5)}{x^5})+\ln(x^5)$

$\Rightarrow 5+\ln(x^5)$, Now evaluating the derivative at $x=1$. We get,

$\Rightarrow [5+\ln(x^5)]|_{x=1}=5+\ln(1)=5$

Approach 2 (Using Logarithmic Properties)

$\frac{d}{dx}(x\ln(x^5))$

$\Rightarrow x\frac{d}{dx}(\ln(x^5))+\ln(x^5)\frac{d}{dx}(x)$

Writing $\ln(x^5)$ as $5\ln(x)$ in the expression, we get,

$\Rightarrow 5x\frac{d}{dx}\ln(x)+5\ln(x)\frac{d}{dx}x$

$\Rightarrow 5+5\ln(x)=5(1+\ln(x))$

Again, evaluating the derivative at $x=1$, we obtain,

$\Rightarrow 5(1+\ln(x))|_{x=1}=5(1+\ln(1))=5$

Happy Differentiating, Cheers