What is the derivative of y=sqrt(ax+b) ?

The derivative of y=sqrt(ax+b) is a/(2sqrt(ax+b))

What is the first derivative of y=sqrt(ax+b) ?

The first derivative of y=sqrt(ax+b) is a/(2sqrt(ax+b))

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derivative of

y = a x + b

\frac{a}{2 a x + b}

Solution steps

\frac{d}{d x} (a x + b)

Apply the chain rule:

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

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

\frac{d}{d x} (a x + b) = a

= \frac{1}{2 a x + b} a

Simplify

\frac{1}{2 a x + b} a : \frac{a}{2 a x + b}

= \frac{a}{2 a x + b}

x^{2} = y, x^{2} = - (y - 2)

\frac{\partial}{\partial x} (x + 3 y^{\frac{1}{3}})

- y^{^{'}} + \frac{3 y}{x} = \frac{y ^{2}}{x}

\frac{d}{d x} (\frac{7 x ^{2} + 3 x + 7}{x})

x \to 0 + lim (7 (\frac{1}{x} - cot (x)))

What is the derivative of y=sqrt(ax+b) ?
The derivative of y=sqrt(ax+b) is a/(2sqrt(ax+b))
What is the first derivative of y=sqrt(ax+b) ?
The first derivative of y=sqrt(ax+b) is a/(2sqrt(ax+b))