What is the derivative of R(t)=(t+e^t)(3-sqrt(t)) ?

The derivative of R(t)=(t+e^t)(3-sqrt(t)) is (-3t-2e^tt+6e^tsqrt(t)-e^t)/(2sqrt(t))+3

What is the first derivative of R(t)=(t+e^t)(3-sqrt(t)) ?

The first derivative of R(t)=(t+e^t)(3-sqrt(t)) is (-3t-2e^tt+6e^tsqrt(t)-e^t)/(2sqrt(t))+3

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derivative of

R (t) = (t + e^{t}) (3 - t)

\frac{- 3 t - 2 e ^{t} t + 6 e ^{t} t - e ^{t}}{2 t} + 3

Solution steps

\frac{d}{d t} ((t + e^{t}) (3 - t))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t + e^{t}, g = 3 - t

= \frac{d}{d t} (t + e^{t}) (3 - t) + \frac{d}{d t} (3 - t) (t + e^{t})

\frac{d}{d t} (t + e^{t}) = 1 + e^{t}

\frac{d}{d t} (3 - t) = - \frac{1}{2 t}

= (1 + e^{t}) (3 - t) + (- \frac{1}{2 t}) (t + e^{t})

Simplify

(1 + e^{t}) (3 - t) + (- \frac{1}{2 t}) (t + e^{t}) : \frac{- 3 t - 2 e ^{t} t + 6 e ^{t} t - e ^{t}}{2 t} + 3

= \frac{- 3 t - 2 e ^{t} t + 6 e ^{t} t - e ^{t}}{2 t} + 3

\frac{\partial}{\partial x} (\frac{x}{x ^{3} - y ^{5}})

n = 0 \sum \infty \frac{7}{n ( n + 3 )}

x = - y^{2} + 6, y = \frac{x}{2} - \frac{3}{2}

y^{^{'}} + 8 y = x

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

What is the derivative of R(t)=(t+e^t)(3-sqrt(t)) ?
The derivative of R(t)=(t+e^t)(3-sqrt(t)) is (-3t-2e^tt+6e^tsqrt(t)-e^t)/(2sqrt(t))+3
What is the first derivative of R(t)=(t+e^t)(3-sqrt(t)) ?
The first derivative of R(t)=(t+e^t)(3-sqrt(t)) is (-3t-2e^tt+6e^tsqrt(t)-e^t)/(2sqrt(t))+3