What is the derivative of y=3cos(x)+7ex+ln(x^2+1) ?

The derivative of y=3cos(x)+7ex+ln(x^2+1) is -3sin(x)+7e+(2x)/(x^2+1)

What is the first derivative of y=3cos(x)+7ex+ln(x^2+1) ?

The first derivative of y=3cos(x)+7ex+ln(x^2+1) is -3sin(x)+7e+(2x)/(x^2+1)

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derivative of

y = 3 cos (x) + 7 e x + ln (x^{2} + 1)

- 3 sin (x) + 7 e + \frac{2 x}{x ^{2} + 1}

Solution steps

\frac{d}{d x} (3 cos (x) + 7 e x + ln (x^{2} + 1))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 cos (x)) + \frac{d}{d x} (7 e x) + \frac{d}{d x} (ln (x^{2} + 1))

\frac{d}{d x} (3 cos (x)) = - 3 sin (x)

\frac{d}{d x} (7 e x) = 7 e

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

= - 3 sin (x) + 7 e + \frac{2 x}{x ^{2} + 1}

f (x) = \frac{4 x - 1}{x}

2^{z}

y = sin (x^{- 1})

x^{2} cos (lo g_{10} (x))

\frac{d}{d t} (2 t - 2)

What is the derivative of y=3cos(x)+7ex+ln(x^2+1) ?
The derivative of y=3cos(x)+7ex+ln(x^2+1) is -3sin(x)+7e+(2x)/(x^2+1)
What is the first derivative of y=3cos(x)+7ex+ln(x^2+1) ?
The first derivative of y=3cos(x)+7ex+ln(x^2+1) is -3sin(x)+7e+(2x)/(x^2+1)