What is the derivative of y=ln(x^5)+e^x-ln(e^2) ?

The derivative of y=ln(x^5)+e^x-ln(e^2) is e^x+5/x

What is the first derivative of y=ln(x^5)+e^x-ln(e^2) ?

The first derivative of y=ln(x^5)+e^x-ln(e^2) is e^x+5/x

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derivative of

y = ln (x^{5}) + e^{x} - ln (e^{2})

e^{x} + \frac{5}{x}

Solution steps

\frac{d}{d x} (ln (x^{5}) + e^{x} - ln (e^{2}))

Simplify

ln (x^{5}) + e^{x} - ln (e^{2}) : ln (x^{5}) + e^{x} - 2

= \frac{d}{d x} (ln (x^{5}) + e^{x} - 2)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (ln (x^{5})) + \frac{d}{d x} (e^{x}) - \frac{d}{d x} (2)

\frac{d}{d x} (ln (x^{5})) = \frac{5}{x}

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

\frac{d}{d x} (2) = 0

= \frac{5}{x} + e^{x} - 0

Simplify

= e^{x} + \frac{5}{x}

y = (3 x + 1)^{2}

\frac{d}{d x} (- y + x + 2)

\int \frac{sec ^{2} ( x )}{2 x} d x

(x - 1) \cdot ln (x - 1)

x \to 0 lim (\frac{x ^{3} + 3 x}{x})

What is the derivative of y=ln(x^5)+e^x-ln(e^2) ?
The derivative of y=ln(x^5)+e^x-ln(e^2) is e^x+5/x
What is the first derivative of y=ln(x^5)+e^x-ln(e^2) ?
The first derivative of y=ln(x^5)+e^x-ln(e^2) is e^x+5/x