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

The derivative of f(x)=(x-ln(1+x))/(x^2) is (-x^2-2x+2xln(x+1)+2ln(x+1))/(x^3(x+1))

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

The first derivative of f(x)=(x-ln(1+x))/(x^2) is (-x^2-2x+2xln(x+1)+2ln(x+1))/(x^3(x+1))

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{2}) = 2 x

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

Simplify

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

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

\int 63 sec^{4} (θ) tan^{6} (θ) d θ

\frac{d}{d x} (x y + ln (\frac{x}{y}))

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{25}{3}

\int 2 y - 5 d y

x \to + 3 - lim (f (x))

What is the derivative of f(x)=(x-ln(1+x))/(x^2) ?
The derivative of f(x)=(x-ln(1+x))/(x^2) is (-x^2-2x+2xln(x+1)+2ln(x+1))/(x^3(x+1))
What is the first derivative of f(x)=(x-ln(1+x))/(x^2) ?
The first derivative of f(x)=(x-ln(1+x))/(x^2) is (-x^2-2x+2xln(x+1)+2ln(x+1))/(x^3(x+1))