To differentiate the function y = ln((1-3x^2)/√(1+3x^2)), we can use the chain rule and the quotient rule.
First, let's simplify the function:
y = ln((1-3x^2)/√(1+3x^2))
= ln((1-3x^2)/(1+3x^2)^(1/2))
Now, let's differentiate using the chain rule:
dy/dx = (d/dx) ln((1-3x^2)/(1+3x^2)^(1/2))
Applying the quotient rule:
dy/dx = [(1+3x^2)^(1/2)(d/dx)(1-3x^2) - (1-3x^2)(d/dx)(1+3x^2)^(1/2)] / [(1+3x^2) (1-3x^2)]
Now, let's find the derivative of each term:
(d/dx) (1-3x^2) = -6x
(d/dx) (1+3x^2)^(1/2) = (1/2)(1+3x^2)^(-1/2)(6x) = 3x/(1+3x^2)^(1/2)
Substituting these results into the expression for dy/dx:
dy/dx = [(1+3x^2)^(1/2)(-6x) - (1-3x^2)(3x/(1+3x^2)^(1/2))] / [(1+3x^2) (1-3x^2)]
Simplifying:
dy/dx = [-6x(1+3x^2) - 3x(1-3x^2)] / [(1+3x^2) (1-3x^2)]
= -9x(1+3x^2) / [(1+3x^2) (1-3x^2)]
= -9x / (1-3x^2)
Therefore, the derivative of y = ln((1-3x^2)/√(1+3x^2)) is dy/dx = -9x / (1-3x^2).
For the function y = ((sin(4x))/(1+cos(x)))^5, we can use the chain rule to differentiate.
Let's denote u = (sin4x) and v = (1+cosx). Then, the function can be written as y = (u/v)^5.
Now, let's find the derivative using the chain rule:
dy/dx = (d/dx) (u/v)^5
= 5(u/v)^4 (d/dx) (u/v)
Using the quotient rule to find the derivative of (u/v):
(d/dx) (u/v) = [(v)(d/dx)(u) - (u)(d/dx)(v)] / v^2
= [(v)(d/dx)(sin4x) - (u)(d/dx)(1+cosx)] / v^2
= [(v)(4cos4x) - (u)(-sinx)] / v^2
= [4vcos4x + usinx] / v^2
Substituting this result into the expression for dy/dx:
dy/dx = 5(u/v)^4 [4vcos4x + usinx] / v^2
= 20(u/v)^4 (vcos4x + (sinx/u)) / v^2
= 20(u/v)^4 (v^2cos4x + sinx) / (v^3)
= 20(cos4x + sinx) / (v)
Since u = sin4x and v = 1+cosx, we can substitute these back in:
dy/dx = 20(cos4x + sinx) / (1+cosx)
Therefore, the derivative of y = ((sin(4x))/(1+cos(x)))^5 is dy/dx = 20(cos4x + sinx) / (1+cosx).
Differentiate the following functions
y=In((1-3x^2)/1+3x^2)^1/2
y=((sin4x)/1+cosx)^5
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