What is the derivative of 1/x+tan(x) ?

The derivative of 1/x+tan(x) is -1/(x^2)+sec^2(x)

What is the first derivative of 1/x+tan(x) ?

The first derivative of 1/x+tan(x) is -1/(x^2)+sec^2(x)

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derivative of

\frac{1}{x} + tan (x)

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

Solution steps

\frac{d}{d x} (\frac{1}{x} + tan (x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (tan (x))

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

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

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

\int \frac{1}{- 8 x - 8 x} d x

f (x) = x^{3} - 4 x, at x = - 2

\int_{2}^{4} \frac{2 x + 5}{5 - x} d x

y^{^{'}} = x e^{- y}

3 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

What is the derivative of 1/x+tan(x) ?
The derivative of 1/x+tan(x) is -1/(x^2)+sec^2(x)
What is the first derivative of 1/x+tan(x) ?
The first derivative of 1/x+tan(x) is -1/(x^2)+sec^2(x)