What is the derivative of f(x)=(2500x)/(ln(8x+8)) ?

The derivative of f(x)=(2500x)/(ln(8x+8)) is (2500(ln(8x+8)(x+1)-x))/(ln^2(8x+8)(x+1))

What is the first derivative of f(x)=(2500x)/(ln(8x+8)) ?

The first derivative of f(x)=(2500x)/(ln(8x+8)) is (2500(ln(8x+8)(x+1)-x))/(ln^2(8x+8)(x+1))

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derivative of

f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

\frac{2500 ( ln ( 8 x + 8 ) ( x + 1 ) - x )}{ln ^{2} ( 8 x + 8 ) ( x + 1 )}

Solution steps

\frac{d}{d x} (\frac{2500 x}{ln ( 8 x + 8 )})

Take the constant out:

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

= 2500 \frac{d}{d x} (\frac{x}{ln ( 8 x + 8 )})

Apply the Quotient Rule:

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

= 2500 \cdot \frac{\frac{d x}{d x} ln ( 8 x + 8 ) - \frac{d}{d x} ( ln ( 8 x + 8 ) ) x}{( ln ( 8 x + 8 ) ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (ln (8 x + 8)) = \frac{1}{x + 1}

= 2500 \cdot \frac{1 \cdot ln ( 8 x + 8 ) - \frac{1}{x + 1} x}{( ln ( 8 x + 8 ) ) ^{2}}

Simplify

2500 \cdot \frac{1 \cdot ln ( 8 x + 8 ) - \frac{1}{x + 1} x}{ln ^{2} ( 8 x + 8 )} : \frac{2500 ( ln ( 8 x + 8 ) ( x + 1 ) - x )}{ln ^{2} ( 8 x + 8 ) ( x + 1 )}

= \frac{2500 ( ln ( 8 x + 8 ) ( x + 1 ) - x )}{ln ^{2} ( 8 x + 8 ) ( x + 1 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

f (x) = x^{2} - x + 4, at x = 1

f (x) = x^{2} - 4, (4, 12)

\frac{d}{d x} (e^{(5 x)})

What is the derivative of f(x)=(2500x)/(ln(8x+8)) ?
The derivative of f(x)=(2500x)/(ln(8x+8)) is (2500(ln(8x+8)(x+1)-x))/(ln^2(8x+8)(x+1))
What is the first derivative of f(x)=(2500x)/(ln(8x+8)) ?
The first derivative of f(x)=(2500x)/(ln(8x+8)) is (2500(ln(8x+8)(x+1)-x))/(ln^2(8x+8)(x+1))