What is the derivative of (4-x^{1/(x-3)}) ?

The derivative of (4-x^{1/(x-3)}) is (-(ln(4-x))/((x-3)^2)-1/((4-x)(x-3)))(4-x)^{1/(x-3)}

What is the first derivative of (4-x^{1/(x-3)}) ?

The first derivative of (4-x^{1/(x-3)}) is (-(ln(4-x))/((x-3)^2)-1/((4-x)(x-3)))(4-x)^{1/(x-3)}

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\frac{d}{d x} ((4 - x)^{\frac{1}{x - 3}})

(- \frac{ln ( 4 - x )}{( x - 3 ) ^{2}} - \frac{1}{( 4 - x ) ( x - 3 )}) (4 - x)^{\frac{1}{x - 3}}

Solution steps

\frac{d}{d x} ((4 - x)^{\frac{1}{x - 3}})

Apply exponent rule:

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

(4 - x)^{\frac{1}{x - 3}} = e^{\frac{1}{x - 3} l n (4 - x)}

= \frac{d}{d x} (e^{\frac{1}{x - 3} l n (4 - x)})

Apply the chain rule:

e^{\frac{1}{x - 3} l n (4 - x)} \frac{d}{d x} (\frac{1}{x - 3} ln (4 - x))

= e^{\frac{1}{x - 3} l n (4 - x)} \frac{d}{d x} (\frac{1}{x - 3} ln (4 - x))

\frac{d}{d x} (\frac{1}{x - 3} ln (4 - x)) = - \frac{ln ( 4 - x )}{( x - 3 ) ^{2}} - \frac{1}{( 4 - x ) ( x - 3 )}

= e^{\frac{1}{x - 3} l n (4 - x)} (- \frac{ln ( 4 - x )}{( x - 3 ) ^{2}} - \frac{1}{( 4 - x ) ( x - 3 )})

e^{\frac{1}{x - 3} l n (4 - x)} = (4 - x)^{\frac{1}{x - 3}}

= (- \frac{ln ( 4 - x )}{( x - 3 ) ^{2}} - \frac{1}{( 4 - x ) ( x - 3 )}) (4 - x)^{\frac{1}{x - 3}}

y = ln (x + 3)

y^{^{''}} + 9 y = - sin (3 t)

y = 2 x^{3} + 3 x^{2} - 4 x, y = x^{2}

y^{^{''}} + 4 y^{^{'}} + 34 y = 0

f (x) = \frac{2}{4 x + 3}

What is the derivative of (4-x^{1/(x-3)}) ?
The derivative of (4-x^{1/(x-3)}) is (-(ln(4-x))/((x-3)^2)-1/((4-x)(x-3)))(4-x)^{1/(x-3)}
What is the first derivative of (4-x^{1/(x-3)}) ?
The first derivative of (4-x^{1/(x-3)}) is (-(ln(4-x))/((x-3)^2)-1/((4-x)(x-3)))(4-x)^{1/(x-3)}