What is the derivative of f(x)=(x^2-7)/(16-x^2) ?

The derivative of f(x)=(x^2-7)/(16-x^2) is (18x)/((16-x^2)^2)

What is the first derivative of f(x)=(x^2-7)/(16-x^2) ?

The first derivative of f(x)=(x^2-7)/(16-x^2) is (18x)/((16-x^2)^2)

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derivative of

f (x) = \frac{x ^{2} - 7}{16 - x ^{2}}

\frac{18 x}{( 16 - x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 7}{16 - x ^{2}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{2} - 7 ) ( 16 - x ^{2} ) - \frac{d}{d x} ( 16 - x ^{2} ) ( x ^{2} - 7 )}{( 16 - x ^{2} ) ^{2}}

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

\frac{d}{d x} (16 - x^{2}) = - 2 x

= \frac{2 x ( 16 - x ^{2} ) - ( - 2 x ) ( x ^{2} - 7 )}{( 16 - x ^{2} ) ^{2}}

Simplify

\frac{2 x ( 16 - x ^{2} ) - ( - 2 x ) ( x ^{2} - 7 )}{( 16 - x ^{2} ) ^{2}} : \frac{18 x}{( 16 - x ^{2} ) ^{2}}

= \frac{18 x}{( 16 - x ^{2} ) ^{2}}

\int lo g_{a} (x) d x

\frac{\partial}{\partial y} (x y^{\frac{1}{2}})

\frac{d}{d x} ((3 - x^{2}) (x^{3} - 5 x + 2))

\frac{d x}{d t} = t x - 2 x + t - 2

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What is the derivative of f(x)=(x^2-7)/(16-x^2) ?
The derivative of f(x)=(x^2-7)/(16-x^2) is (18x)/((16-x^2)^2)
What is the first derivative of f(x)=(x^2-7)/(16-x^2) ?
The first derivative of f(x)=(x^2-7)/(16-x^2) is (18x)/((16-x^2)^2)