What is the derivative of (t+e^t)(6-sqrt(t)) ?

The derivative of (t+e^t)(6-sqrt(t)) is (-3t-2e^tt+12e^tsqrt(t)-e^t)/(2sqrt(t))+6

What is the first derivative of (t+e^t)(6-sqrt(t)) ?

The first derivative of (t+e^t)(6-sqrt(t)) is (-3t-2e^tt+12e^tsqrt(t)-e^t)/(2sqrt(t))+6

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derivative of

(t + e^{t}) (6 - t)

\frac{- 3 t - 2 e ^{t} t + 12 e ^{t} t - e ^{t}}{2 t} + 6

Solution steps

\frac{d}{d t} ((t + e^{t}) (6 - t))

Apply the Product Rule:

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

f = t + e^{t}, g = 6 - t

= \frac{d}{d t} (t + e^{t}) (6 - t) + \frac{d}{d t} (6 - t) (t + e^{t})

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

\frac{d}{d t} (6 - t) = - \frac{1}{2 t}

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

Simplify

(1 + e^{t}) (6 - t) + (- \frac{1}{2 t}) (t + e^{t}) : \frac{- 3 t - 2 e ^{t} t + 12 e ^{t} t - e ^{t}}{2 t} + 6

= \frac{- 3 t - 2 e ^{t} t + 12 e ^{t} t - e ^{t}}{2 t} + 6

y^{^{''}} + 5 y^{^{'}} + 6 y = 6

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

y = e^{- s i n (x)}

q^{^{'}} + \frac{q}{20 + t} = 1

y^{^{'}} = x^{2} \frac{1}{y}

What is the derivative of (t+e^t)(6-sqrt(t)) ?
The derivative of (t+e^t)(6-sqrt(t)) is (-3t-2e^tt+12e^tsqrt(t)-e^t)/(2sqrt(t))+6
What is the first derivative of (t+e^t)(6-sqrt(t)) ?
The first derivative of (t+e^t)(6-sqrt(t)) is (-3t-2e^tt+12e^tsqrt(t)-e^t)/(2sqrt(t))+6