What is the derivative of h(-x)=(x^2+9)/(10) ?

The derivative of h(-x)=(x^2+9)/(10) is x/5

What is the first derivative of h(-x)=(x^2+9)/(10) ?

The first derivative of h(-x)=(x^2+9)/(10) is x/5

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derivative of

h (- x) = \frac{x ^{2} + 9}{10}

\frac{x}{5}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 9}{10})

Take the constant out:

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

= \frac{1}{10} \frac{d}{d x} (x^{2} + 9)

Apply the Sum/Difference Rule:

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

= \frac{1}{10} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))

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

\frac{d}{d x} (9) = 0

= \frac{1}{10} (2 x + 0)

Simplify

\frac{1}{10} (2 x + 0) : \frac{x}{5}

= \frac{x}{5}

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

\int e^{7 x} sin (6 x) d x

y = \frac{1}{t} + \frac{1}{t ^{2}} - \frac{1}{t}

(x^{2} + a^{2})^{e^{x^{2}}}

\frac{d}{d x} (2 a - b x)

What is the derivative of h(-x)=(x^2+9)/(10) ?
The derivative of h(-x)=(x^2+9)/(10) is x/5
What is the first derivative of h(-x)=(x^2+9)/(10) ?
The first derivative of h(-x)=(x^2+9)/(10) is x/5