Consider this function a composition of two functions, f(g(x)). In this case, \(\displaystyle f(x) =2 \sin (x)\) and \(\displaystyle g(x) = x^2\). According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x))\cdot g'(x)\). Here, \(\displaystyle f'(g(x)) = 2 \cos (x^2 )\) and \(\displaystyle g'(x) = 2x\), so the derivative is \(\displaystyle 2 \cos (x^2 ) \cdot 2x = 4x \cdot \cos(x^2)\)

According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x))\cdot g'(x)\). In this case, \(\displaystyle f(x) = \cos(x)\) and \(\displaystyle g(x) = \ln(x)\). The derivative is \(\displaystyle - \sin (\ln x ) \cdot \frac{1}{x } = \frac{- \sin (\ln x )}{x }\).

According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x))\cdot g'(x)\). In this case, \(\displaystyle f(x) = x^3\) and \(\displaystyle g(x) = \tan x\). The derivative is \(\displaystyle 3 (\tan x) ^2 \cdot( 1 + \tan ^2 x ) = 3 \tan ^2 x (1 + \tan ^2 x ) = 3 \tan ^2 x + 3 \tan ^4 x\)

According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x)) \cdot g'(x)\). In this case, \(\displaystyle f(x ) = 4x^2\) and \(\displaystyle g(x) = x-2\). \(\displaystyle f'(x) = 8x\) and \(\displaystyle g'(x) = 1\). 

The derivative is \(\displaystyle 8(x-2) \cdot 1 = 8x -16\)

According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x)) \cdot g'(x)\). In this case, \(\displaystyle f(x) = e^x\) and \(\displaystyle g(x) = \sin x\). Since \(\displaystyle f'(x) = e^x\) and \(\displaystyle g'(x) = \cos x\), the derivative is \(\displaystyle e ^ {\sin x } \cdot \cos x\)

According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x)) \cdot g'(x)\). In this case, \(\displaystyle f(x) = 3^x\) and \(\displaystyle g(x) = x^2\). Since \(\displaystyle f'(x) = 3^ {x} \cdot \ln(3)\) and \(\displaystyle g'(x) = 2x\), the derivative is \(\displaystyle 3^{x^2 } \cdot \ln(3 ) \cdot 2x\)

According to the chain rule, \(\displaystyle f(g(x)) ' = f'(g(x)) \cdot g'(x)\). In this case, \(\displaystyle f(x) = \frac{1}{4} \cdot 2^x\) and \(\displaystyle g(x) = 4x-1\). Here \(\displaystyle f'(x) = \frac{1}{4} \cdot 2^x \cdot \ln(2)\) and \(\displaystyle g'(x) = 4\). The derivative is: 

\(\displaystyle \frac{1}{4} \cdot 2 ^{4x-1} \cdot \ln(2) \cdot 4 = 2^{4x-1} \cdot \ln(2)\)

We begin by taking the derivative of both sides of the equation.

\(\displaystyle \frac{d}{dx}(y^2)=\frac{d}{dx}(x^2+y)\).

\(\displaystyle 2yy' = 2x +y'\). (The left hand side uses the Chain Rule.)

\(\displaystyle 2yy' -y' = 2x\).

\(\displaystyle (2y-1)y'=2x\)

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

\(\displaystyle \frac{dy}{dx} = \frac{2x}{2y-1}\).

We can use implicit differentiation to find \(\displaystyle y'\). We being by taking the derivative of both sides of the equation.

\(\displaystyle \frac{d}{dx}(x^2+xy^2)=\frac{d}{dx}(y)\)

\(\displaystyle \frac{d}{dx}(x^2)+\frac{d}{dx}(xy^2)=y'\)

\(\displaystyle 2x +[x \times 2y' + y^2 \times 1] = y'\) (This line uses the product rule for the derivative of \(\displaystyle xy^2\).)

\(\displaystyle 2x +2xy' +y^2 = y'\)

\(\displaystyle 2x+y^2 = -2xy' + y'\)

\(\displaystyle 2x+y^2 = (-2x+1)y'\)

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

Since we have a function inside of a another function, the chain rule is appropriate here.

The chain rule formula is

\(\displaystyle (h(g(x)))' = h'(g(x))\times g'(x)\).

In our function, both \(\displaystyle h(x), g(x)\) are \(\displaystyle e^x\)

So we have

\(\displaystyle f'(x) = (h(g(x)))' = e^{e^x} \times e^x = e^{e^x+1}\)

and

\(\displaystyle f'(1)=e^{e+1}\).