1) To differentiate the function y = ln((1-3x^2)/(1+3x^2))^(1/2):
Let's simplify the function step by step before differentiating:
1. Apply the power rule for logarithms by bringing the exponent (1/2) down:
y = (1/2)ln((1-3x^2)/(1+3x^2))
2. Apply the chain rule by differentiating the inside function ((1-3x^2)/(1+3x^2)):
y' = (1/2) * 1 / ((1-3x^2)/(1+3x^2)) * ((1+3x^2)/(1-3x^2))'
3. Simplify:
y' = (1/2) * (1+3x^2) / (1-3x^2) * [(1-3x^2)'(1+3x^2) - (1-3x^2)(1+3x^2)'] / (1-3x^2)^2
= (1/2) * (1+3x^2) / (1-3x^2) * [-6x(1+3x^2) - (1-3x^2)6x] / (1-3x^2)^2
= (1/2) * (1+3x^2) / (1-3x^2) * [-6x(1+3x^2) - 6x(1-3x^2)] / (1-3x^2)^2
= (1/2) * (1+3x^2) / (1-3x^2) * (-12x) / (1-3x^2)^2
= -6x(1+3x^2) / [2(1-3x^2)^2]
Therefore, the derivative of y = ln((1-3x^2)/(1+3x^2))^(1/2) is y' = -6x(1+3x^2) / [2(1-3x^2)^2].
2) To differentiate the function y = ((sin4x)/1+cosx)^5:
Let's simplify the function step by step before differentiating:
1. Apply the power rule:
y = (sin4x/(1+cosx))^5
2. Apply the chain rule by differentiating the inside function (sin4x/(1+cosx)):
y' = 5(sin4x / (1+cosx))^4 * (cosx(1+cosx) - sin4x(-sinx)) / (1+cosx)^2
= 5(sin4x)^4 * (cosx(1+cosx) + sin4x(sin4x)) / (1+cosx)^2
= 5(sin^4(4x)) * (cosx(1+cosx) + sin^2(4x)) / (1+cosx)^2
Therefore, the derivative of y = ((sin4x)/1+cosx)^5 is y' = 5(sin^4(4x)) * (cosx(1+cosx) + sin^2(4x)) / (1+cosx)^2.
Differentiate the following functions
y=In((1-3x^2)/(1+3x^2))^1/2
y=((sin4x)/1+cosx)^5
3 answers
y = In((1-3x^2)/(1+3x^2))^1/2 = 1/2 ln((1-3x^2)/(1+3x^2))
= 1/2 [ln(1-3x^2) - ln(1+3x^2)]
y' = 1/2 [ -6x/(1-3x^2) - 6x/(1+3x^2)] = 6x/(9x^4 - 1)
= 1/2 [ln(1-3x^2) - ln(1+3x^2)]
y' = 1/2 [ -6x/(1-3x^2) - 6x/(1+3x^2)] = 6x/(9x^4 - 1)
Apologies for the confusion in my previous response. You are correct.
The correct differentiation of y = ln((1-3x^2)/(1+3x^2))^1/2 is:
y = (1/2)ln((1-3x^2)/(1+3x^2))
Applying the chain rule:
y' = (1/2) * 1/((1-3x^2)/(1+3x^2)) * ((1-3x^2)/(1+3x^2))'
Simplifying:
y' = (1/2) * (1+3x^2)/(1-3x^2) * [-(1-3x^2)'/(1+3x^2) + (1-3x^2)/(1+3x^2)']
Using the quotient rule:
y' = (1/2) * (1+3x^2)/(1-3x^2) * [(-(-6x))/(1+3x^2) - (1-3x^2)(6x)/(1+3x^2)^2]
Simplifying further:
y' = (1/2) * (1+3x^2)/(1-3x^2) * [(6x/(1+3x^2)) - (6x(1-3x^2))/(1+3x^2)^2]
y' = (1/2) * (1+3x^2)/(1-3x^2) * [(6x - 6x + 18x^3)/(1+3x^2)^2]
y' = (1/2) * (1+3x^2)/(1-3x^2) * (18x^3)/(1+3x^2)^2
simplifying, we have:
y' = 9x^3/(1-9x^4)
The correct differentiation of y = ln((1-3x^2)/(1+3x^2))^1/2 is:
y = (1/2)ln((1-3x^2)/(1+3x^2))
Applying the chain rule:
y' = (1/2) * 1/((1-3x^2)/(1+3x^2)) * ((1-3x^2)/(1+3x^2))'
Simplifying:
y' = (1/2) * (1+3x^2)/(1-3x^2) * [-(1-3x^2)'/(1+3x^2) + (1-3x^2)/(1+3x^2)']
Using the quotient rule:
y' = (1/2) * (1+3x^2)/(1-3x^2) * [(-(-6x))/(1+3x^2) - (1-3x^2)(6x)/(1+3x^2)^2]
Simplifying further:
y' = (1/2) * (1+3x^2)/(1-3x^2) * [(6x/(1+3x^2)) - (6x(1-3x^2))/(1+3x^2)^2]
y' = (1/2) * (1+3x^2)/(1-3x^2) * [(6x - 6x + 18x^3)/(1+3x^2)^2]
y' = (1/2) * (1+3x^2)/(1-3x^2) * (18x^3)/(1+3x^2)^2
simplifying, we have:
y' = 9x^3/(1-9x^4)