What is the derivative of f(x)=(c^4-x^4)/(c^4+x^4) ?

The derivative of f(x)=(c^4-x^4)/(c^4+x^4) is -(8c^4x^3)/((c^4+x^4)^2)

What is the first derivative of f(x)=(c^4-x^4)/(c^4+x^4) ?

The first derivative of f(x)=(c^4-x^4)/(c^4+x^4) is -(8c^4x^3)/((c^4+x^4)^2)

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derivative of

f (x) = \frac{c ^{4} - x ^{4}}{c ^{4} + x ^{4}}

- \frac{8 c ^{4} x ^{3}}{( c ^{4} + x ^{4} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{c ^{4} - x ^{4}}{c ^{4} + x ^{4}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( c ^{4} - x ^{4} ) ( c ^{4} + x ^{4} ) - \frac{d}{d x} ( c ^{4} + x ^{4} ) ( c ^{4} - x ^{4} )}{( c ^{4} + x ^{4} ) ^{2}}

\frac{d}{d x} (c^{4} - x^{4}) = - 4 x^{3}

\frac{d}{d x} (c^{4} + x^{4}) = 4 x^{3}

= \frac{( - 4 x ^{3} ) ( c ^{4} + x ^{4} ) - 4 x ^{3} ( c ^{4} - x ^{4} )}{( c ^{4} + x ^{4} ) ^{2}}

Simplify

\frac{( - 4 x ^{3} ) ( c ^{4} + x ^{4} ) - 4 x ^{3} ( c ^{4} - x ^{4} )}{( c ^{4} + x ^{4} ) ^{2}} : - \frac{8 c ^{4} x ^{3}}{( c ^{4} + x ^{4} ) ^{2}}

= - \frac{8 c ^{4} x ^{3}}{( c ^{4} + x ^{4} ) ^{2}}

\frac{d}{d y} (y^{- 2})

\int_{- 1}^{1} e^{x} - x^{2} + 1 d x

\int \frac{x ^{2}}{x ^{2} + x ^{4}} d x

\int_{0}^{1} 5 cos (\frac{π t}{2}) d t

y^{^{''}} - 4 y^{^{'}} = 16 t

What is the derivative of f(x)=(c^4-x^4)/(c^4+x^4) ?
The derivative of f(x)=(c^4-x^4)/(c^4+x^4) is -(8c^4x^3)/((c^4+x^4)^2)
What is the first derivative of f(x)=(c^4-x^4)/(c^4+x^4) ?
The first derivative of f(x)=(c^4-x^4)/(c^4+x^4) is -(8c^4x^3)/((c^4+x^4)^2)