Answer :
Sure! Let's solve the problem step by step.
### Part 1: Determine the derivative of the function
The function provided is:
[tex]\[ f(x) = x^5 + 35 \][/tex]
To find the derivative, [tex]\( f'(x) \)[/tex], we'll use the power rule for differentiation. The power rule states that if [tex]\( f(x) = x^n \)[/tex], then [tex]\( f'(x) = n \cdot x^{n-1} \)[/tex].
1. Differentiate [tex]\( x^5 \)[/tex]:
[tex]\[ \frac{d}{dx}(x^5) = 5x^4 \][/tex]
2. Differentiate 35, which is a constant:
[tex]\[ \frac{d}{dx}(35) = 0 \][/tex]
Combining these, we get:
[tex]\[ f'(x) = 5x^4 + 0 = 5x^4 \][/tex]
So, the correct choice for the derivative is:
[tex]\[ a. \, f'(x) = 5x^4 \][/tex]
### Part 2: Integrate the given integral
Next, we need to solve the integral:
[tex]\[ \int \frac{\ln|x|}{x^2} \, dx \][/tex]
This integral requires the use of integration techniques. Let's break it down:
1. The integrand is:
[tex]\[ \frac{\ln|x|}{x^2} \][/tex]
2. To integrate this function, we can use the property of the natural logarithm and the power of [tex]\( x \)[/tex].
Let [tex]\( u = \ln|x| \)[/tex], hence [tex]\( du = \frac{1}{x} dx \)[/tex].
Now, rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int \frac{u}{x} \cdot \frac{1}{x} dx = \int u \cdot du \][/tex]
3. Integrate [tex]\( u \)[/tex]:
[tex]\[ \int u \, du = \frac{u^2}{2} = \frac{(\ln|x|)^2}{2} \][/tex]
So, the antiderivative of [tex]\( \frac{\ln|x|}{x^2} \)[/tex] is:
[tex]\[ \frac{(\ln|x|)^2}{-2} \][/tex]
Thus, the integral evaluates to:
[tex]\[ \int \frac{\ln|x|}{x^2} \, dx = -\frac{(\ln|x|)^2}{2} + C \][/tex]
Here, [tex]\( C \)[/tex] represents the constant of integration.
### Summary
To summarize, we have:
1. The derivative of [tex]\( f(x) = x^5 + 35 \)[/tex] is: [tex]\( f'(x) = 5x^4 \)[/tex], which matches option a.
2. The integral [tex]\( \int \frac{\ln|x|}{x^2} \, dx \)[/tex] evaluates to [tex]\( -\frac{(\ln|x|)^2}{2} + C \)[/tex].
I hope this detailed explanation helps you understand how to find the derivative and solve the integral!
### Part 1: Determine the derivative of the function
The function provided is:
[tex]\[ f(x) = x^5 + 35 \][/tex]
To find the derivative, [tex]\( f'(x) \)[/tex], we'll use the power rule for differentiation. The power rule states that if [tex]\( f(x) = x^n \)[/tex], then [tex]\( f'(x) = n \cdot x^{n-1} \)[/tex].
1. Differentiate [tex]\( x^5 \)[/tex]:
[tex]\[ \frac{d}{dx}(x^5) = 5x^4 \][/tex]
2. Differentiate 35, which is a constant:
[tex]\[ \frac{d}{dx}(35) = 0 \][/tex]
Combining these, we get:
[tex]\[ f'(x) = 5x^4 + 0 = 5x^4 \][/tex]
So, the correct choice for the derivative is:
[tex]\[ a. \, f'(x) = 5x^4 \][/tex]
### Part 2: Integrate the given integral
Next, we need to solve the integral:
[tex]\[ \int \frac{\ln|x|}{x^2} \, dx \][/tex]
This integral requires the use of integration techniques. Let's break it down:
1. The integrand is:
[tex]\[ \frac{\ln|x|}{x^2} \][/tex]
2. To integrate this function, we can use the property of the natural logarithm and the power of [tex]\( x \)[/tex].
Let [tex]\( u = \ln|x| \)[/tex], hence [tex]\( du = \frac{1}{x} dx \)[/tex].
Now, rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int \frac{u}{x} \cdot \frac{1}{x} dx = \int u \cdot du \][/tex]
3. Integrate [tex]\( u \)[/tex]:
[tex]\[ \int u \, du = \frac{u^2}{2} = \frac{(\ln|x|)^2}{2} \][/tex]
So, the antiderivative of [tex]\( \frac{\ln|x|}{x^2} \)[/tex] is:
[tex]\[ \frac{(\ln|x|)^2}{-2} \][/tex]
Thus, the integral evaluates to:
[tex]\[ \int \frac{\ln|x|}{x^2} \, dx = -\frac{(\ln|x|)^2}{2} + C \][/tex]
Here, [tex]\( C \)[/tex] represents the constant of integration.
### Summary
To summarize, we have:
1. The derivative of [tex]\( f(x) = x^5 + 35 \)[/tex] is: [tex]\( f'(x) = 5x^4 \)[/tex], which matches option a.
2. The integral [tex]\( \int \frac{\ln|x|}{x^2} \, dx \)[/tex] evaluates to [tex]\( -\frac{(\ln|x|)^2}{2} + C \)[/tex].
I hope this detailed explanation helps you understand how to find the derivative and solve the integral!
