What is the derivative of a(x+1(x-r)) ?

The derivative of a(x+1(x-r)) is a(2x-x(dr)/(dx)-r-(dr)/(dx)+1)

What is the first derivative of a(x+1(x-r)) ?

The first derivative of a(x+1(x-r)) is a(2x-x(dr)/(dx)-r-(dr)/(dx)+1)

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\frac{d}{d x} (a (x + 1) (x - r))

a (2 x - x \frac{d r}{d x} - r - \frac{d r}{d x} + 1)

Solution steps

\frac{d}{d x} (a (x + 1) (x - r))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= a \frac{d}{d x} ((x + 1) (x - r))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x + 1, g = x - r

= a (\frac{d}{d x} (x + 1) (x - r) + \frac{d}{d x} (x - r) (x + 1))

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

\frac{d}{d x} (x - r) = 1 - \frac{d r}{d x}

= a (1 \cdot (x - r) + (1 - \frac{d r}{d x}) (x + 1))

Simplify

a (1 \cdot (x - r) + (1 - \frac{d r}{d x}) (x + 1)) : a (2 x - x \frac{d r}{d x} - r - \frac{d r}{d x} + 1)

= a (2 x - x \frac{d r}{d x} - r - \frac{d r}{d x} + 1)

t y^{^{'}} + y = 0

\frac{d x}{d t} = x - 4

\int_{1}^{\infty} \frac{x}{x ^{4} + 1} d x

y = \frac{1 + 2 x}{3 + x}, (2, 1)

\frac{\partial}{\partial x} (5 x^{2} + 4 y^{2})

What is the derivative of a(x+1(x-r)) ?
The derivative of a(x+1(x-r)) is a(2x-x(dr)/(dx)-r-(dr)/(dx)+1)
What is the first derivative of a(x+1(x-r)) ?
The first derivative of a(x+1(x-r)) is a(2x-x(dr)/(dx)-r-(dr)/(dx)+1)