What is the derivative of f(x)=(c^5-x^5)/(c^5+c^5) ?

The derivative of f(x)=(c^5-x^5)/(c^5+c^5) is -(5x^4)/(2c^5)

What is the first derivative of f(x)=(c^5-x^5)/(c^5+c^5) ?

The first derivative of f(x)=(c^5-x^5)/(c^5+c^5) is -(5x^4)/(2c^5)

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derivative of

f (x) = \frac{c ^{5} - x ^{5}}{c ^{5} + c ^{5}}

- \frac{5 x ^{4}}{2 c ^{5}}

Solution steps

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

Take the constant out:

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

= \frac{1}{c ^{5} + c ^{5}} \frac{d}{d x} (c^{5} - x^{5})

Apply the Sum/Difference Rule:

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

= \frac{1}{c ^{5} + c ^{5}} (\frac{d}{d x} (c^{5}) - \frac{d}{d x} (x^{5}))

\frac{d}{d x} (c^{5}) = 0

\frac{d}{d x} (x^{5}) = 5 x^{4}

= \frac{1}{c ^{5} + c ^{5}} (0 - 5 x^{4})

Simplify

\frac{1}{c ^{5} + c ^{5}} (0 - 5 x^{4}) : - \frac{5 x ^{4}}{2 c ^{5}}

= - \frac{5 x ^{4}}{2 c ^{5}}

f^{^{'}} (t) - f (t) = 6 t

y^{2} + x^{2} y^{^{'}} = x y y^{^{'}}

\int 8 sin^{2} (t - \frac{π}{12}) d t

2 sin^{2} (t)

f (x) = \frac{x}{e ^{x} - 1}

What is the derivative of f(x)=(c^5-x^5)/(c^5+c^5) ?
The derivative of f(x)=(c^5-x^5)/(c^5+c^5) is -(5x^4)/(2c^5)
What is the first derivative of f(x)=(c^5-x^5)/(c^5+c^5) ?
The first derivative of f(x)=(c^5-x^5)/(c^5+c^5) is -(5x^4)/(2c^5)