What is the derivative of (t+e^t)(9-sqrt(t)) ?

The derivative of (t+e^t)(9-sqrt(t)) is (-3t-2e^tt+18e^tsqrt(t)-e^t)/(2sqrt(t))+9

What is the first derivative of (t+e^t)(9-sqrt(t)) ?

The first derivative of (t+e^t)(9-sqrt(t)) is (-3t-2e^tt+18e^tsqrt(t)-e^t)/(2sqrt(t))+9

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derivative of

(t + e^{t}) (9 - t)

\frac{- 3 t - 2 e ^{t} t + 18 e ^{t} t - e ^{t}}{2 t} + 9

Solution steps

\frac{d}{d t} ((t + e^{t}) (9 - t))

Apply the Product Rule:

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

f = t + e^{t}, g = 9 - t

= \frac{d}{d t} (t + e^{t}) (9 - t) + \frac{d}{d t} (9 - t) (t + e^{t})

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

\frac{d}{d t} (9 - t) = - \frac{1}{2 t}

= (1 + e^{t}) (9 - t) + (- \frac{1}{2 t}) (t + e^{t})

Simplify

(1 + e^{t}) (9 - t) + (- \frac{1}{2 t}) (t + e^{t}) : \frac{- 3 t - 2 e ^{t} t + 18 e ^{t} t - e ^{t}}{2 t} + 9

= \frac{- 3 t - 2 e ^{t} t + 18 e ^{t} t - e ^{t}}{2 t} + 9

3 y^{^{'''}} + 12 y^{^{''}} + 12 y^{^{'}} = 0

\int_{0}^{1} (x^{2} + x) e^{x} d x

\int_{0}^{1} e^{- t^{2}} d t

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 0

sec^{2} (2 x + 5) \cdot 2

What is the derivative of (t+e^t)(9-sqrt(t)) ?
The derivative of (t+e^t)(9-sqrt(t)) is (-3t-2e^tt+18e^tsqrt(t)-e^t)/(2sqrt(t))+9
What is the first derivative of (t+e^t)(9-sqrt(t)) ?
The first derivative of (t+e^t)(9-sqrt(t)) is (-3t-2e^tt+18e^tsqrt(t)-e^t)/(2sqrt(t))+9