What is the derivative of f(x)=4^{c/x} ?

The derivative of f(x)=4^{c/x} is -(ln(2)*2^{(x+2c)/x}c)/(x^2)

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

The first derivative of f(x)=4^{c/x} is -(ln(2)*2^{(x+2c)/x}c)/(x^2)

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derivative of

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

- \frac{ln ( 2 ) \cdot 2 ^{\frac{x + 2 c}{x}} c}{x ^{2}}

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

4^{\frac{c}{x}} = e^{\frac{c}{x} l n (4)}

= \frac{d}{d x} (e^{\frac{c}{x} l n (4)})

Apply the chain rule:

e^{\frac{c}{x} l n (4)} \frac{d}{d x} (\frac{c}{x} ln (4))

= e^{\frac{c}{x} l n (4)} \frac{d}{d x} (\frac{c}{x} ln (4))

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

= e^{\frac{c}{x} l n (4)} (- \frac{2 ln ( 2 ) c}{x ^{2}})

Simplify

e^{\frac{c}{x} l n (4)} (- \frac{2 ln ( 2 ) c}{x ^{2}}) : - \frac{ln ( 2 ) \cdot 2 ^{\frac{x + 2 c}{x}} c}{x ^{2}}

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

\frac{d}{d x} (\frac{1}{4} sin^{4} (x))

x \to \infty lim (1 + 2^{x})

y = x^{2} - 1, (5, 24)

\frac{d}{d x} (x^{5} \cdot ln (x))

\frac{d}{d x} (x^{2} - a^{2})

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