Differentiate each function
a) y = Cos³ x
b) y = Sin(x³)
c) y = Sin²xCos3x
a)
let z = cos x then dz/dx = -sin x
y = z^3
y' = 3 z^2 dz/dx
y' = 3 cos^2 x (-sin x)
= -3 sin x cos^2 x
b)
let z = x^3 then dz/dx = 3 x^2
y = sin z
y' = cos z dz/dx
= cos(x^3) (3 x^2)
= 3 x^2 cos x^3
c)y' = sin^2 x (-3sin 3x)+ cos 3x (2 sin x cos x
= -3sin 3x sin^2 x + 2 cos 3x sin x cos x
To differentiate the given functions, we can use the chain rule and the product rule of differentiation. Let's differentiate each function step by step:
a) y = cos³x
To differentiate y = cos³x, we need to apply the chain rule. The derivative of cos³x is given by:
dy/dx = d/dx (cos³x) = 3cos²x * (-sinx)
So, the derivative of y = cos³x is dy/dx = -3cos²x * sinx.
b) y = sin(x³)
To differentiate y = sin(x³), we also need to apply the chain rule. The derivative of sin(x³) is given by:
dy/dx = d/dx (sin(x³)) = 3x²cos(x³)
So, the derivative of y = sin(x³) is dy/dx = 3x²cos(x³).
c) y = sin²x*cos(3x)
To differentiate y = sin²x * cos(3x), we need to 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:
d(uv)/dx = u * dv/dx + v * du/dx
In this case, u(x) = sin²x and v(x) = cos(3x). Let's differentiate it step by step:
du/dx = d/dx (sin²x) = 2sinx * cosx
dv/dx = d/dx (cos(3x)) = -3sin(3x)
Now, we can use the product rule to find the derivative of y = sin²x * cos(3x):
dy/dx = sin²x * (-3sin(3x)) + cos(3x) * (2sinx * cosx)
Simplifying further, we get:
dy/dx = -3sin³(3x) + 2sinx * cosx * cos(3x)
Therefore, the derivative of y = sin²x * cos(3x) is dy/dx = -3sin³(3x) + 2sinx * cosx * cos(3x).