differentiate each function
a) y = cos^3x
b) y = sin(x^3)
c) y = sin^2 xcos3x
To differentiate each function, we can use the chain rule and the product rule of differentiation. Let's go through the steps for each function:
a) y = cos^3x:
Step 1: Recall that the chain rule states that if we have a composition of functions, say f(g(x)), the derivative of f(g(x)) with respect to x is given by f'(g(x)) multiplied by g'(x).
In this case, our outer function is f(u) = u^3, and our inner function is g(x) = cos(x).
Step 2: Find the derivative of the outer function: f'(u) = 3u^2.
Step 3: Find the derivative of the inner function: g'(x) = -sin(x).
Step 4: Apply the chain rule by multiplying the derivatives found in steps 2 and 3:
dy/dx = f'(g(x)) * g'(x) = 3(cos^2(x)) * (-sin(x)).
Hence, the derivative of y = cos^3x is dy/dx = -3cos^2(x)sin(x).
b) y = sin(x^3):
Step 1: Apply the chain rule again. Our outer function is f(u) = sin(u), and our inner function is g(x) = x^3.
Step 2: Find the derivative of the outer function: f'(u) = cos(u).
Step 3: Find the derivative of the inner function: g'(x) = 3x^2.
Step 4: Multiply the derivatives from steps 2 and 3 to apply the chain rule:
dy/dx = f'(g(x)) * g'(x) = cos(x^3) * (3x^2).
So, the derivative of y = sin(x^3) is dy/dx = 3x^2cos(x^3).
c) y = sin^2(x)cos(3x):
Step 1: In this case, we have two functions multiplied together. We'll need to apply the product rule of differentiation.
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'v + uv').
Step 2: Let's assign u(x) = sin^2(x) and v(x) = cos(3x).
Step 3: Find the derivatives of u(x) and v(x):
u'(x) = 2sin(x)cos(x) (using the chain rule),
v'(x) = -3sin(3x) (using the chain rule).
Step 4: Apply the product rule:
dy/dx = u'(x)v(x) + u(x)v'(x) = 2sin(x)cos(x) * cos(3x) + sin^2(x) * (-3sin(3x)).
Therefore, the derivative of y = sin^2(x)cos(3x) is dy/dx = 2sin(x)cos(x)cos(3x) - 3sin^3(x)sin(3x).