To find the derivative of this function we must use the Chain Rule and the Quotient Rule. Applying the Chain Rule to the numerator gives

\(\displaystyle h'(x) = -\sin(\sin(x))\cos(x)\)

Now using the Quotient Rule for the function, we find the derivative to be

\(\displaystyle f'(x) = \frac{[-\sin(\sin(x))\cos(x)]\sin(x)-\cos(x)\cos(\sin(x))}{\sin^2(x)}\)

To find the derivative of this function, we must use the Quotient Rule which is 

\(\displaystyle f'(x) = \frac{h'(x)g(x) - h(x)g'(x)}{(g(x))^2}\)

Applying this to the function we are given, with \(\displaystyle h(x) = x^2\) and \(\displaystyle g(x) = \sin(x)\) gives us 

\(\displaystyle f'(x) = \frac{2x\sin(x)-x^2\cos(x)}{\sin^2(x)}\)

To find the derivative of this function we must use the Product Rule and the Quotient Rule. Appling the Product Rule to the numerator of the function gives us

\(\displaystyle h'(x) = \sin(x)+x\cos(x)\)

Using this with the Quotient Rule, we find 

\(\displaystyle f'(x) = \frac{[\sin(x)+x\cos(x)]\cos(x)-x\sin(x)(-\sin(x))}{\cos^2(x)} \rightarrow\)

\(\displaystyle f'(x) = \frac{\sin(x)\cos(x)+x\cos^2(x)+x\sin^2(x)}{\cos^2(x)}\)

To find the derivative of this function we must use the Product Rule and the Chain Rule. If \(\displaystyle h(x) = \sin(x)\) and \(\displaystyle g(x) = \cos(2x)\) then

\(\displaystyle h'(x) = \cos(x)\)

\(\displaystyle g'(x) = -\sin(2x)(2) = -2\sin(2x)\)

Applying these derivatives to the Product Rule gives us

\(\displaystyle f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)\)

To find the derivative of this function we must use the Product Rule and the Chain Rule. First we set 

\(\displaystyle h(x) = \frac{1}{x}\)

and

\(\displaystyle g(x) = \cos^2(x)\)

Now differentiating both of these functions gives

\(\displaystyle h'(x) = \frac{-1}{x^2}\)

\(\displaystyle g'(x) = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x)\)

Applying this to the Product Rule gives us,

\(\displaystyle f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)\)

To find the derivative of this function we must use the Product Rule and the Chain Rule. If we have

\(\displaystyle h(x) = \sin(x)\) and \(\displaystyle g(x) = \sin\left(\frac{1}{x}\right)\) then

\(\displaystyle h'(x) = \cos(x)\) and \(\displaystyle g'(x) = \cos\left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)\)

Applying this to the product rule, we find

\(\displaystyle f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)+\cos\left(\frac{1}{x}\right)\sin(x)\left(\frac{-1}{x^2}\right) \rightarrow\)

\(\displaystyle f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}\)

To find the derivative of this function, we must use the Product Rule, Quotient Rule, and the Chain Rule. To do this, we first find the derivative of each part.

\(\displaystyle h(x) = x^4 \Rightarrow h'(x) = 4x^3\)

\(\displaystyle g(x) = \cos(3x) \Rightarrow g'(x) = -3\sin(3x)\)

\(\displaystyle j(x) = \sin(x) \Rightarrow j'(x) = \cos(x)\)

Using the derivatives of each part we can find the derivative of the numerator using the Product Rule

\(\displaystyle h(x)g(x) = x^4\cos(3x) \Rightarrow \\ h'(x)g(x) + h(x)g'(x) = 4x^3\cos(3x)+(-3x^4)\sin(3x)\)

Finally, putting this into the Quotient Rule gives

\(\displaystyle f'(x) = \frac{[4x^3\cos(3x)+(-3x^4)\sin(3x)]\sin(x)-\cos(x)x^4\cos(3x)}{\sin^2x}\)

\(\displaystyle f'(x) = \frac{4x^3\cos(3x)\sin(x)-3x^4\sin(3x)\sin(x)-x^4\cos(3x)\cos(x)}{\sin^2x}\)

To differentiate this function we must use the Quotient Rule

\(\displaystyle f'(x) = \frac{h'(x)g(x)-h(x)g'(x)}{(g(x))^2}\)

Using \(\displaystyle h(x) = \sin(x)\) and \(\displaystyle g(x) = x^2\).

The derivative of the function is then

\(\displaystyle f'(x) = \frac{x^2\cos(x)-2x\sin(x)}{x^4}\)

To differentiate this function we must use the Chain Rule. where

\(\displaystyle f(x) = h(g(x)) \Rightarrow f'(x) = h'(g(x))\cdot g'(x)\)

Therefore the derivative of the function is

\(\displaystyle f(x) = 2\sin(\cos(2x))\cos(\cos(2x))(-\sin(2x))(2) \rightarrow\)

\(\displaystyle f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)\)

To find the derivative of the function, we must use the Product Rule, 

\(\displaystyle f'(x) = h'(x)g(x)+h(x)g'(x)\)

Using \(\displaystyle h(x) = x^2\) and \(\displaystyle g(x) = \cos(x)\).

Therefore

\(\displaystyle f'(x) = 2x\cos(x) + x^2(-\sin(x)) = 2x\cos(x) -x^2\sin(x)\)

\(\displaystyle f'(x) = 2x\cos(x) - x^2\sin(x)\)