What is the derivative of f(x)=x^2+(4(1-x)^2)/pi ?

The derivative of f(x)=x^2+(4(1-x)^2)/pi is ((8+2pi)x-8)/pi

What is the first derivative of f(x)=x^2+(4(1-x)^2)/pi ?

The first derivative of f(x)=x^2+(4(1-x)^2)/pi is ((8+2pi)x-8)/pi

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derivative of

f (x) = x^{2} + \frac{4 ( 1 - x ) ^{2}}{π}

\frac{( 8 + 2 π ) x - 8}{π}

Solution steps

\frac{d}{d x} (x^{2} + \frac{4 ( 1 - x ) ^{2}}{π})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (\frac{4 ( 1 - x ) ^{2}}{π})

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (\frac{4 ( 1 - x ) ^{2}}{π}) = - \frac{8}{π} (1 - x)

= 2 x - \frac{8}{π} (1 - x)

Simplify

2 x - \frac{8}{π} (1 - x) : \frac{( 8 + 2 π ) x - 8}{π}

= \frac{( 8 + 2 π ) x - 8}{π}

\int 4 ln (9 x^{2}) d x

\int z ln (2)

\int e^{l n (x)} d x

\frac{d y}{d x} = 2 y + 1 cos (x)

(\frac{( x + 3 )}{( x - 1 )})^{2}

What is the derivative of f(x)=x^2+(4(1-x)^2)/pi ?
The derivative of f(x)=x^2+(4(1-x)^2)/pi is ((8+2pi)x-8)/pi
What is the first derivative of f(x)=x^2+(4(1-x)^2)/pi ?
The first derivative of f(x)=x^2+(4(1-x)^2)/pi is ((8+2pi)x-8)/pi