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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\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)

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

y = arccos (\frac{11}{x})

\int x e^{x} y d x

\int_{1}^{l n (x)} \frac{1}{t} d t

\int \frac{1}{x ^{\frac{2}{3}}} d x

\int \frac{x ^{3}}{( x - 2 ) ( x + 3 )} d x

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