What is the derivative of y=e^4+ln(x) ?

The derivative of y=e^4+ln(x) is 1/x

What is the first derivative of y=e^4+ln(x) ?

The first derivative of y=e^4+ln(x) is 1/x

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derivative of

y = e^{4} + ln (x)

\frac{1}{x}

Solution steps

\frac{d}{d x} (e^{4} + ln (x))

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (e^{4}) = 0

\frac{d}{d x} (ln (x)) = \frac{1}{x}

= 0 + \frac{1}{x}

Simplify

= \frac{1}{x}

3 sin^{2} (x) + 2 tan (x)

(2 x^{2} + 3 x + 2)^{5}

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

1 + sin^{2} (x)

2 e^{3} x

What is the derivative of y=e^4+ln(x) ?
The derivative of y=e^4+ln(x) is 1/x
What is the first derivative of y=e^4+ln(x) ?
The first derivative of y=e^4+ln(x) is 1/x