What is the derivative of 2x^2+4x-5 ?

The derivative of 2x^2+4x-5 is 4x+4

What is the first derivative of 2x^2+4x-5 ?

The first derivative of 2x^2+4x-5 is 4x+4

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derivative of

2 x^{2} + 4 x - 5

4 x + 4

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{2}) + \frac{d}{d x} (4 x) - \frac{d}{d x} (5)

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

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (5) = 0

= 4 x + 4 - 0

Simplify

= 4 x + 4

n = 0 \sum \infty 3 \cdot (- \frac{1}{5})^{n}

t^{2} y^{^{''}} + 19 t y^{^{'}} + 85 y = 0

\int - \frac{1}{64} e^{- 4 x} d x

y^{^{'}} = e^{2 y} sin (3 x)

y^{^{'}} = \frac{3 sec ( y )}{( x + 1 ) ^{2}}

What is the derivative of 2x^2+4x-5 ?
The derivative of 2x^2+4x-5 is 4x+4
What is the first derivative of 2x^2+4x-5 ?
The first derivative of 2x^2+4x-5 is 4x+4