What is the derivative of h(x)=(2/x-3)(x^2+7) ?

The derivative of h(x)=(2/x-3)(x^2+7) is -(2(x^2+7))/(x^2)+4-6x

What is the first derivative of h(x)=(2/x-3)(x^2+7) ?

The first derivative of h(x)=(2/x-3)(x^2+7) is -(2(x^2+7))/(x^2)+4-6x

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derivative of

h (x) = (\frac{2}{x} - 3) (x^{2} + 7)

- \frac{2 ( x ^{2} + 7 )}{x ^{2}} + 4 - 6 x

Solution steps

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

Apply the Product Rule:

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

f = \frac{2}{x} - 3, g = x^{2} + 7

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

\frac{d}{d x} (\frac{2}{x} - 3) = - \frac{2}{x ^{2}}

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

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

Simplify

(- \frac{2}{x ^{2}}) (x^{2} + 7) + 2 x (\frac{2}{x} - 3) : - \frac{2 ( x ^{2} + 7 )}{x ^{2}} + 4 - 6 x

= - \frac{2 ( x ^{2} + 7 )}{x ^{2}} + 4 - 6 x

x \to 0 lim (ln (\frac{1 + x}{x}))

\int (x^{2} cos (x)) d x

\int sin (\frac{n \cdot π}{3} x) d x

\int \frac{y}{1 + x y} d x

y = 1 - x^{2}, x = 2

What is the derivative of h(x)=(2/x-3)(x^2+7) ?
The derivative of h(x)=(2/x-3)(x^2+7) is -(2(x^2+7))/(x^2)+4-6x
What is the first derivative of h(x)=(2/x-3)(x^2+7) ?
The first derivative of h(x)=(2/x-3)(x^2+7) is -(2(x^2+7))/(x^2)+4-6x