What is the derivative of f(x)=sqrt(x)+40 ?

The derivative of f(x)=sqrt(x)+40 is 1/(2sqrt(x))

What is the first derivative of f(x)=sqrt(x)+40 ?

The first derivative of f(x)=sqrt(x)+40 is 1/(2sqrt(x))

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derivative of

f (x) = x + 40

\frac{1}{2 x}

Solution steps

\frac{d}{d x} (x + 40)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x) + \frac{d}{d x} (40)

\frac{d}{d x} (x) = \frac{1}{2 x}

\frac{d}{d x} (40) = 0

= \frac{1}{2 x} + 0

Simplify

= \frac{1}{2 x}

n \to \infty lim (1 + \frac{1}{n})

y^{^{'}} + 2 t y = 2 t

y = 7 + 5 x^{2} - 2 x^{3}

x, x^{3}

- \frac{12 x}{( 6 x ^{2} + 5 ) ^{2}}

What is the derivative of f(x)=sqrt(x)+40 ?
The derivative of f(x)=sqrt(x)+40 is 1/(2sqrt(x))
What is the first derivative of f(x)=sqrt(x)+40 ?
The first derivative of f(x)=sqrt(x)+40 is 1/(2sqrt(x))