What is the derivative of f(x)=(cx^2)/(x^2+b) ?

The derivative of f(x)=(cx^2)/(x^2+b) is (2cbx)/((x^2+b)^2)

What is the first derivative of f(x)=(cx^2)/(x^2+b) ?

The first derivative of f(x)=(cx^2)/(x^2+b) is (2cbx)/((x^2+b)^2)

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derivative of

f (x) = \frac{c x ^{2}}{x ^{2} + b}

\frac{2 c b x}{( x ^{2} + b ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{c x ^{2}}{x ^{2} + b})

Take the constant out:

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

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

Apply the Quotient Rule:

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

= c \frac{\frac{d}{d x} ( x ^{2} ) ( x ^{2} + b ) - \frac{d}{d x} ( x ^{2} + b ) x ^{2}}{( x ^{2} + b ) ^{2}}

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

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

= c \frac{2 x ( x ^{2} + b ) - 2 x x ^{2}}{( x ^{2} + b ) ^{2}}

Simplify

c \frac{2 x ( x ^{2} + b ) - 2 x x ^{2}}{( x ^{2} + b ) ^{2}} : \frac{2 c b x}{( x ^{2} + b ) ^{2}}

= \frac{2 c b x}{( x ^{2} + b ) ^{2}}

\frac{\partial}{\partial x} (5 x + y (x^{2} + y^{2}))

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

y = ln (7 - 9 x^{2} + 8 x^{4}), at x = 1

y = x^{2} + 3, (- 3, 12)

n = 1 \sum \infty \frac{1}{( - 3 ) ^{n}}

What is the derivative of f(x)=(cx^2)/(x^2+b) ?
The derivative of f(x)=(cx^2)/(x^2+b) is (2cbx)/((x^2+b)^2)
What is the first derivative of f(x)=(cx^2)/(x^2+b) ?
The first derivative of f(x)=(cx^2)/(x^2+b) is (2cbx)/((x^2+b)^2)