If f(x)=3 sin x+ln(5x), find f '(x).
f(x)=3 sin x+ln(5x)
f'(x) = 3 cosx + 1/x
Note that ln(5x) = ln5 + lnx
So, the ln5 is just a constant, with zero derivative
Of course, you could use the chain rule, so
d/dx ln(5x) = (1/5x)*5 = 5/5x = 1/x
To find the derivative of f(x) with respect to x, we will use the sum rule and the chain rule.
Using the sum rule, we take the derivative of each term in the function separately:
The derivative of 3 sin(x) is 3 cos(x).
The derivative of ln(5x) is 1 / (5x) * 5 = 1 / x.
Therefore, the derivative of f(x) = 3 sin(x) + ln(5x) is:
f '(x) = 3 cos(x) + 1 / x.
To find the derivative of the function f(x) = 3sin(x) + ln(5x), we can apply the rules of differentiation. Let's break it down step by step.
Step 1: Differentiate the term 3sin(x).
To differentiate sin(x), we can use the chain rule. The derivative of sin(x) is cos(x). Since we have a constant coefficient of 3, the derivative of 3sin(x) is 3cos(x).
Step 2: Differentiate the term ln(5x).
To differentiate ln(5x), we can use the chain rule and the derivative of ln(u), which is 1/u. The derivative of ln(5x) is (1/(5x)) * d(5x)/dx.
Step 3: Differentiate the function using the sum rule.
The derivative of f(x) = 3sin(x) + ln(5x) is the sum of the derivatives of the individual terms. So, f'(x) = 3cos(x) + (1/(5x)) * 5.
Simplifying this equation gives:
f'(x) = 3cos(x) + 1/x
Therefore, the derivative of the function f(x) = 3sin(x) + ln(5x) is f'(x) = 3cos(x) + 1/x.