Rules of Logarithms Functions Question

Question

I cannot understand this question in my textbook

a) Show carefully how to use the rules of logarithms to deduce that ln(x^(3x+1)) = (3x + 1) ln(x) whenever x > 0. State which rule(s) you are using in general form

(b) Giving full details of all steps, show how to use the product rule for derivatives to find the derivative of (3x + 1) ln(x).

(c) Show how to find the derivative of x^(3x+1) by first writing it in a different exponential form and then using the results of parts (a) and (b) above. State which rules you use along the way

Answer 1

Show carefully how to use the rules of logarithms to deduce that ln(x^3x+1) = (3x + 1)*ln(x) whenever x > 0. State which rule(s) you are using in general form

1. Recall that Natural Logarithm Product Rule is given by ln(x*y) = ln(x) + ln(y)
2. ln(x^3x+1) is the same as ln(x*x*x...*x) for 3x+1 factors of x
3. Using the Natural Logarithm Product Rule —ln(x^3x+1) = ln(x) + ln(x) + ln(x) ... + ln(x) for 3x+1 terms
4. Therefore, ln(x^3x+1) = (3x + 1)*ln(x)

Giving full details of all steps, show how to use the Product Rule for Derivatives to find the derivative of (3x + 1)*ln(x).

1. Recall the Product Rule for Derivatives is given by d/dx[f(x)*g(x)] = f'(x)*g(x) + f(x)*g'(x)
2. Let f(x) = 3x + 1 and let g(x) = ln(x)
3. Using the Power Rule for Derivatives — d/dx[xⁿ] = nx^n-1 and the Constant Rule for Derivatives — d/dx[c] = 0, evaluate the derivative of f(x)
4. f(x) = 3x + 1 —> f'(x) = 3
5. Recall the definition of the derivative of a natural logarithm — d/dx[ln(x)] = 1/x, and evaluate the derivative of g(x)
6. g(x) = ln(x) —> g'(x) = 1/x
7. Substitute calculated values into the general formula for the Product Rule for Derivatives to find the derivative of (3x + 1)*ln(x)
8. d/dx[f(x)*g(x)] = f'(x)*g(x) + f(x)*g'(x)
9. d/dx[(3x + 1)*ln(x)] = (3)*(ln(x)) + (3x + 1)*(1/x)
10. d/dx[(3x + 1)*ln(x)] = (3*ln(x)) + ((3x + 1)/x)
11. Therefore the derivative is 3ln(x) + (3x+1)/x

Show how to find the derivative of x^3x+1 by first writing it in a different exponential form and then using the results of parts above. State which rules you use along the way

1. Utilize the rule e^ln(x) = x
2. x^3x+1 = e^ln(x^3x+1)
3. x^3x+1 = e^ln(x^3x*x¹⁾
4. Next, utilize the Natural Logarithm Product Rule ln(x*y) = ln(x) + ln(y)
5. x^3x+1 = e^ln(x^{3x) + ln(x)}
6. x^3x+1 = e^{3x*ln(x) + ln(x)}
7. x^3x+1 = e^ln(x)*(3x+1)
8. Recall that d/du[e^u] = e^u du
9. Let u = ln(x)*(3x+1)
10. We have already derived this exact formula in the previous part
11. Therefore, we know du = 3ln(x) + (3x+1)/x
12. Finally, we can evaluate d/du[e^u] = e^u du
13. d/dx[e^ln(x)*(3x+1)] = e^ln(x)*(3x+1) * (3ln(x) + (3x+1)/x)
14. Therefore, the derivative of x^3x+1 is given by e^ln(x)*(3x+1) * (3ln(x) + (3x+1)/x)

Answer 2

The answer to your questions

a) It uses the power logarithm rule, which is derived from the product logarithm rule.

If you have an xⁿ you can right it as a product of n-number of x's.

log xⁿ= log (x.x.x.......(n-th x))= log ∏_n x =log x+ log x+ ...... (nth) log x = n . log x

so, log (x^(3x+1)) = (3x+1).log x

b) d/dx ((3x+1). log x) =(3x+1). d/dx (log x) + log x d/dx (3x+1)

= (3x+1)* 1/x + (log x )* (d/dx (3x) + d/dx 1)

= 3 + 1/x + (log x)*3

= 3+ 1/x + 3* log x

c) Let, y= x^(3x+1) , now, d/dx log y = d/dx log (x^(3x+1))

differentiating both sides and using answers from a, we get

1/y * d/dx y = d/dx ((3x+1). log x)

Left side looks like the step b. Let's use that

1/y * d/dx y = 3+ 1/x +3* log x

d/dx y = y * (3+ 1/x +3* log x)

So, we can say, d/dx (x^(3x+1)) = x^(3x+1) * (3+ 1/x +3* log x)

Answer 3