What is the derivative of e^e+2ln(1+x^2-3^x) ?

The derivative of e^e+2ln(1+x^2-3^x) is (4x)/(x^2+1)-ln(3)*3^x

What is the first derivative of e^e+2ln(1+x^2-3^x) ?

The first derivative of e^e+2ln(1+x^2-3^x) is (4x)/(x^2+1)-ln(3)*3^x

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Popular >

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (e^{e}) = 0

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

\frac{d}{d x} (3^{x}) = 3^{x} ln (3)

= 0 + \frac{4 x}{1 + x ^{2}} - 3^{x} ln (3)

Simplify

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

\int \frac{3 x}{6 x ^{2} + 9} d x

f (x) = - \frac{2 x - 10}{x - 3}, at x = 2

y = x^{2}, x = y^{2}, y = 1

\frac{d}{d x} (\frac{x ^{3}}{x ^{3}})

\int \frac{1}{w} d w

What is the derivative of e^e+2ln(1+x^2-3^x) ?
The derivative of e^e+2ln(1+x^2-3^x) is (4x)/(x^2+1)-ln(3)*3^x
What is the first derivative of e^e+2ln(1+x^2-3^x) ?
The first derivative of e^e+2ln(1+x^2-3^x) is (4x)/(x^2+1)-ln(3)*3^x