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Algebra of Derivatives

If y=ex+e-x...

Question

If $y = e^{x} + e^{- x} + log x^{2}$ , find $\frac{d y}{d x}$ .

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Solution

$y = e^{x} + e^{- x} + log x^{2}$

Differentiating w.r.t. x, we get,

$\frac{d y}{d x} = e^{x} - e^{- x} + \frac{2 x}{x^{2}}$

$\frac{d y}{d x} = e^{x} - e^{- x} + \frac{2}{x}$

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Similar questions

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

\frac{d y}{d x} = \frac{l o g x}{(1 - l o g x)^{2}}

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x^{y} = e^{(x - y)}

\frac{d y}{d x} = \frac{log x}{(1 + log x)^{2}}

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

\frac{d y}{d x} = \frac{log x}{(1 + log x)^{2}}

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

then prove that:

\frac{d y}{d x} = \frac{log x}{{(1 + log x)}^{2}}

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, then show that

\frac{d y}{d x} = \frac{log x}{(1 + log x)^{2}}

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