What is the derivative of ((x+1^3)/3+1/(4x+4)) ?

The derivative of ((x+1^3)/3+1/(4x+4)) is (4x^4+16x^3+20x^2+8x-1)/(4(x+1)^2)+1

What is the first derivative of ((x+1^3)/3+1/(4x+4)) ?

The first derivative of ((x+1^3)/3+1/(4x+4)) is (4x^4+16x^3+20x^2+8x-1)/(4(x+1)^2)+1

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\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

\frac{4 x ^{4} + 16 x ^{3} + 20 x ^{2} + 8 x - 1}{4 ( x + 1 ) ^{2}} + 1

Solution steps

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3}) + \frac{d}{d x} (\frac{1}{4 x + 4})

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

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

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

Simplify

(x + 1)^{2} - \frac{1}{4 ( x + 1 ) ^{2}} : \frac{4 x ^{4} + 16 x ^{3} + 20 x ^{2} + 8 x - 1}{4 ( x + 1 ) ^{2}} + 1

= \frac{4 x ^{4} + 16 x ^{3} + 20 x ^{2} + 8 x - 1}{4 ( x + 1 ) ^{2}} + 1

3 x^{2} e^{x}

\int 2 sin (x) + x d x

\frac{d}{d x} (5 x + \frac{3}{x})

x \to 0 lim (sin (x) ln (3 x))

(\frac{1}{4} x^{8} - 3 e^{x} + 100 ln (x))^{^{'}}

What is the derivative of ((x+1^3)/3+1/(4x+4)) ?
The derivative of ((x+1^3)/3+1/(4x+4)) is (4x^4+16x^3+20x^2+8x-1)/(4(x+1)^2)+1
What is the first derivative of ((x+1^3)/3+1/(4x+4)) ?
The first derivative of ((x+1^3)/3+1/(4x+4)) is (4x^4+16x^3+20x^2+8x-1)/(4(x+1)^2)+1