# differentiate f(p)=ln[(p+1)^2(p+2)^3(p+3)^4]?

### 4 Answers

- 2 weeks agoFavorite Answer
Break it apart

ln((p + 1)^(2) * (p + 2)^(3) * (p + 3)^(4)) =>

ln((p + 1)^2) + ln((p + 2)^3) + ln((p + 3)^4) =>

2 * ln(p + 1) + 3 * ln(p + 2) + 4 * ln(p + 3)

Now it's easy to differentiate.

f'(p) = 2/(p + 1) + 3/(p + 2) + 4/(p + 3)

There you go. You can combine it all, if you want, but this looks fine to me.

- MyRankLv 62 weeks ago
f(p) = ln [(p + 1)² (p + 2)³ (p + 3)⁴]

f’(p) = 1/ (p + 1)² (p + 2)³ (p + 3)⁴ d/dp ((p + 1)² (p + 2)³ (p + 3)⁴)(or)f(p) = log (p + 1)² + log (p + 2)³ + log (p + 3)⁴f’(p) = 1/ (p + 1)² (2(p + 1)) + 1/(p + 2)³ (3(p + 2)²) + 1/(p + 3)⁴ (4(p + 3)³)= 2/ (p + 1) + 3/ (p + 2) + 4/(p + 3).

Source(s): https://myrank.co.in/ - Jim MoorLv 72 weeks ago
Find the derivative of the following via implicit differentiation:

d/dp(f(p)) = d/dp(log((1 + p)^2 (2 + p)^3 (3 + p)^4))

Using the chain rule, d/dp(f(p)) = ( df(u))/( du) ( du)/( dp), where u = p and d/( du)(f(u)) = f'(u):

(d/dp(p)) f'(p) = d/dp(log((1 + p)^2 (2 + p)^3 (3 + p)^4))

The derivative of p is 1:

1 f'(p) = d/dp(log((1 + p)^2 (2 + p)^3 (3 + p)^4))

Simplify log((p + 1)^2 (p + 2)^3 (p + 3)^4) using the identity log(a b) = log(a) + log(b):

f'(p) = d/dp(log((1 + p)^2) + log((2 + p)^3) + log((3 + p)^4))

Differentiate the sum term by term:

f'(p) = d/dp(log((1 + p)^2)) + d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4))

Simplify log((p + 1)^2) using the identity log(a^b) = b log(a):

f'(p) = d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4)) + d/dp(2 log(1 + p))

Factor out constants:

f'(p) = d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4)) + 2 (d/dp(log(1 + p)))

Using the chain rule, d/dp(log(p + 1)) = ( dlog(u))/( du) ( du)/( dp), where u = p + 1 and d/( du)(log(u)) = 1/u:

f'(p) = d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4)) + 2 (d/dp(1 + p))/(1 + p)

Differentiate the sum term by term:

f'(p) = d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4)) + d/dp(1) + d/dp(p) 2/(1 + p)

The derivative of 1 is zero:

f'(p) = d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4)) + (2 (d/dp(p) + 0))/(1 + p)

Simplify the expression:

f'(p) = (2 (d/dp(p)))/(1 + p) + d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4))

The derivative of p is 1:

f'(p) = d/dp(log((2 + p)^3)) + d/dp(log((3 + p)^4)) + 1 2/(1 + p)

Simplify log((p + 2)^3) using the identity log(a^b) = b log(a):

f'(p) = 2/(1 + p) + d/dp(log((3 + p)^4)) + d/dp(3 log(2 + p))

Factor out constants:

f'(p) = 2/(1 + p) + d/dp(log((3 + p)^4)) + 3 (d/dp(log(2 + p)))

Using the chain rule, d/dp(log(p + 2)) = ( dlog(u))/( du) ( du)/( dp), where u = p + 2 and d/( du)(log(u)) = 1/u:

f'(p) = 2/(1 + p) + d/dp(log((3 + p)^4)) + 3 (d/dp(2 + p))/(2 + p)

Differentiate the sum term by term:

f'(p) = 2/(1 + p) + d/dp(log((3 + p)^4)) + d/dp(2) + d/dp(p) 3/(2 + p)

The derivative of 2 is zero:

f'(p) = 2/(1 + p) + d/dp(log((3 + p)^4)) + (3 (d/dp(p) + 0))/(2 + p)

Simplify the expression:

f'(p) = 2/(1 + p) + (3 (d/dp(p)))/(2 + p) + d/dp(log((3 + p)^4))

The derivative of p is 1:

f'(p) = 2/(1 + p) + d/dp(log((3 + p)^4)) + 1 3/(2 + p)

Simplify log((p + 3)^4) using the identity log(a^b) = b log(a):

f'(p) = 2/(1 + p) + 3/(2 + p) + d/dp(4 log(3 + p))

Factor out constants:

f'(p) = 2/(1 + p) + 3/(2 + p) + 4 (d/dp(log(3 + p)))

Using the chain rule, d/dp(log(p + 3)) = ( dlog(u))/( du) ( du)/( dp), where u = p + 3 and d/( du)(log(u)) = 1/u:

f'(p) = 2/(1 + p) + 3/(2 + p) + 4 (d/dp(3 + p))/(3 + p)

Differentiate the sum term by term:

f'(p) = 2/(1 + p) + 3/(2 + p) + d/dp(3) + d/dp(p) 4/(3 + p)

The derivative of 3 is zero:

f'(p) = 2/(1 + p) + 3/(2 + p) + (4 (d/dp(p) + 0))/(3 + p)

Simplify the expression:

f'(p) = 2/(1 + p) + 3/(2 + p) + (4 (d/dp(p)))/(3 + p)

The derivative of p is 1:

Answer: |

| f'(p) = 2/(1 + p) + 3/(2 + p) + 1 4/(3 + p)

thanks. im new to this so i do apprecite it.