d/dx[lnsquare root of(x^3) cube root of (x+3) fifth root of (3x-2)]
are we taking ln of the whole thing??
I read it as
d( ln [ (x^3)^(1/2) (x+3)^(1/3) (3x-2)^(1/5) ] /dx
let y = ln [ (x^3)^(1/2) (x+3)^(1/3) (3x-2)^(1/5) ]
remember ln(AB) = lnA + lnB
so our
y = ln (x^3)^(1/2) + ln (x+3)^(1/3) + ln (3x-2)^(1/5)
and remember ln A^n = n lnA
= 3(1/2) lnx + (1/3)ln(x+3) + (1/5)ln(3x-2)
dy/dx = (3/2)(1/x) + (1/3)(1/(x+3) + 1/5)(3/(3x-2))
= 3/(2x) + 1/(3(x+3)) + 3/(5(3x-2))
I would not take it any further, it would look even messier, having to form a common denominator of
30(x+3)(3x-2)
To differentiate the given expression, we can rely on the chain rule. Here's the step-by-step solution:
Step 1: Rewrite the expression using logarithm properties.
f(x) = ln[(x^3)^(1/2) * (x+3)^(1/3) * (3x-2)^(1/5)]
Step 2: Apply the logarithm rule to simplify the expression further.
f(x) = ln(x^(3/2)) + ln((x+3)^(1/3)) + ln((3x-2)^(1/5))
Step 3: Apply the power rule of logarithms to rewrite the expression.
f(x) = (3/2)ln(x) + (1/3)ln(x+3) + (1/5)ln(3x-2)
Step 4: Differentiate each term using the chain rule.
f'(x) = (3/2) * (1/x) + (1/3) * (1/(x+3)) + (1/5) * (1/(3x-2)) * 3
Simplifying further gives:
f'(x) = 3/(2x) + 1/(3x+9) + 3/(5(3x-2))
Thus, the derivative of ln[sqrt(x^3) * (x+3)^(1/3) * (3x-2)^(1/5)] with respect to x is given by 3/(2x) + 1/(3x+9) + 3/(5(3x-2)).
To find the derivative of the given function, which is the product of multiple functions, we will use the product rule. The product rule states that if we have two functions, u(x) and v(x), then the derivative of their product is given by:
(u(x) · v'(x)) + (v(x) · u'(x))
Let's break down the given function into individual parts to make it easier to apply the product rule:
f(x) = ln(sqrt(x^3)) * cbrt(x+3) * (3x-2)^(1/5)
Breaking it down into individual functions:
u(x) = ln(sqrt(x^3))
v(x) = cbrt(x+3)
w(x) = (3x-2)^(1/5)
To find the derivative, we need to find the derivative of each individual function and then apply the product rule.
First, let's find the derivative of each function.
1. Derivative of u(x):
Using the chain rule, we have:
u'(x) = (1/(sqrt(x^3))) * (1/2) * (3x^2)
= (3x^2) / (2sqrt(x^3))
2. Derivative of v(x):
Using the chain rule, we have:
v'(x) = (1/3) * (x+3)^(-2/3)
= (x+3)^(-2/3) / 3
3. Derivative of w(x):
Using the power rule, we have:
w'(x) = (1/5) * (3x-2)^((1/5) - 1) * 3
= (3/5) * (3x-2)^(-4/5)
Now, we can apply the product rule:
f'(x) = u(x) * v'(x) + v(x) * u'(x) + u(x) * w'(x)
Substituting the values we found, we get:
f'(x) = ln(sqrt(x^3)) * ((x+3)^(-2/3)/3) + cbrt(x+3) * ((3x^2) / (2sqrt(x^3))) + ln(sqrt(x^3)) * ((3/5) * (3x-2)^(-4/5))
Simplifying further if needed, the derivative of the given function f(x) will be f'(x) using the above formula.