What is the derivative of f(x)=(2x+1)^5+10x(2x+1)^4 ?

The derivative of f(x)=(2x+1)^5+10x(2x+1)^4 is 960x^4+1600x^3+960x^2+240x+20

What is the first derivative of f(x)=(2x+1)^5+10x(2x+1)^4 ?

The first derivative of f(x)=(2x+1)^5+10x(2x+1)^4 is 960x^4+1600x^3+960x^2+240x+20

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derivative of

f (x) = (2 x + 1)^{5} + 10 x (2 x + 1)^{4}

960 x^{4} + 1600 x^{3} + 960 x^{2} + 240 x + 20

Solution steps

\frac{d}{d x} ((2 x + 1)^{5} + 10 x (2 x + 1)^{4})

Simplify

(2 x + 1)^{5} + 10 x (2 x + 1)^{4} : 192 x^{5} + 400 x^{4} + 320 x^{3} + 120 x^{2} + 20 x + 1

= \frac{d}{d x} (192 x^{5} + 400 x^{4} + 320 x^{3} + 120 x^{2} + 20 x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (192 x^{5}) + \frac{d}{d x} (400 x^{4}) + \frac{d}{d x} (320 x^{3}) + \frac{d}{d x} (120 x^{2}) + \frac{d}{d x} (20 x) + \frac{d}{d x} (1)

\frac{d}{d x} (192 x^{5}) = 960 x^{4}

\frac{d}{d x} (400 x^{4}) = 1600 x^{3}

\frac{d}{d x} (320 x^{3}) = 960 x^{2}

\frac{d}{d x} (120 x^{2}) = 240 x

\frac{d}{d x} (20 x) = 20

\frac{d}{d x} (1) = 0

= 960 x^{4} + 1600 x^{3} + 960 x^{2} + 240 x + 20 + 0

Simplify

= 960 x^{4} + 1600 x^{3} + 960 x^{2} + 240 x + 20

y^{^{'}} + \frac{y ^{2}}{x ^{2}} = 0

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

\frac{d y}{d x} + 4 y = - x

\int e^{2 x} (2) d x

y^{^{'}} = y^{\frac{2}{3}}

What is the derivative of f(x)=(2x+1)^5+10x(2x+1)^4 ?
The derivative of f(x)=(2x+1)^5+10x(2x+1)^4 is 960x^4+1600x^3+960x^2+240x+20
What is the first derivative of f(x)=(2x+1)^5+10x(2x+1)^4 ?
The first derivative of f(x)=(2x+1)^5+10x(2x+1)^4 is 960x^4+1600x^3+960x^2+240x+20