What is the derivative of 2x+1/(ln(x^2+e+1)-4) ?

The derivative of 2x+1/(ln(x^2+e+1)-4) is -(2x)/((x^2+e)(ln(x^2+e)+1)^2)+2

What is the first derivative of 2x+1/(ln(x^2+e+1)-4) ?

The first derivative of 2x+1/(ln(x^2+e+1)-4) is -(2x)/((x^2+e)(ln(x^2+e)+1)^2)+2

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\frac{d}{d x} (2 x + \frac{1}{ln ( x ^{2} + e ) + 1} - 4)

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

Solution steps

\frac{d}{d x} (2 x + \frac{1}{ln ( x ^{2} + e ) + 1} - 4)

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (4) = 0

= 2 - \frac{2 x}{( ln ( x ^{2} + e ) + 1 ) ^{2} ( x ^{2} + e )} - 0

Simplify

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

\frac{x}{4} arcsin (x)

\frac{\partial}{\partial x} (\frac{1}{2 x ^{2}})

x \to 0 lim (\frac{10}{x})

\int_{l n (9)}^{l n (36)} e^{\frac{x}{2}} d x

What is the derivative of 2x+1/(ln(x^2+e+1)-4) ?
The derivative of 2x+1/(ln(x^2+e+1)-4) is -(2x)/((x^2+e)(ln(x^2+e)+1)^2)+2
What is the first derivative of 2x+1/(ln(x^2+e+1)-4) ?
The first derivative of 2x+1/(ln(x^2+e+1)-4) is -(2x)/((x^2+e)(ln(x^2+e)+1)^2)+2