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Calculus/derivative of a function

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Expert: - 2/27/2010


Question
find the derivative of the following
part(a):
y= sqrt(x^4 + 1)tan2x



part(b):
f(x)= sqrt(3x/x^2+4)

part(c):
y= sin(cos(sinx))

Answer
Find the derivatives:

y = √(x^4 + 1)*tan(2x)

Product rule with f(x)= √(x^4 + 1) and g(x) = tan(2x).

The derivative is f(x)g'(x) + f'(x)g(x).

It can be seen that f'(x) = 4x^3/√(x^4 + 1) and g'(x) = 2sec²(2x).
Take f(x), g(x), f'(x), and g'(x) and put them in.


f(x)= √(3x/x^2+4)
Quotient rule with g(x) = 3x and h(x) = x²+4 inside a power rule.

Since f(x) is (g(x)/h(x))^0., the power rule says f'(x) = (0.5/√[3x/(x²+4)])d(g(x)/h(x))dx.

The quotient rule says that d(3x/(x^2+4))dx = [h(x)g'(x) - g(x)h'(x)]/h²(x).

The functions g(x) and h(x) are already knwon.
It can be seen that g'(x) = 3 and h'(x) = 2x.

When g(x), h(x), g'(x), and h'(x) are put in, the result is f'(x) in terms of x alone.


y = sin(cos(sinx))

This a chain rule.  We have sin(f(x)), where f(x) = cos(g(x)), where g(x) = sin(x).

We know the is y = sin(f(x)), dy/dx = cos(f(x))f'(x).
Since f(x) = cos(g(x)), f'(x) = -sin(g(x))g'(x).
Since g(x) = sin(x), g'(x) = cos(x).

Going back up the line to the top, cos(x) can be put in for g'(x) in the line that defines f'(x).
The function g(x) can be put in the definition for f(x) and f'(x).
Now that f(x) and f'(x) are in terms of x, they can both be substituted into dy/dx.

This will give dy/dx in terms of trig functions with x alone.

Calculus

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