What is the derivative of 5*2^{sqrt(t-2)} ?

The derivative of 5*2^{sqrt(t-2)} is (5ln(2)*2^{sqrt(t-2)-1})/(sqrt(t-2))

What is the first derivative of 5*2^{sqrt(t-2)} ?

The first derivative of 5*2^{sqrt(t-2)} is (5ln(2)*2^{sqrt(t-2)-1})/(sqrt(t-2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

5 \cdot 2^{t - 2}

\frac{5 ln ( 2 ) \cdot 2 ^{t - 2 - 1}}{t - 2}

Solution steps

\frac{d}{d t} (5 \cdot 2^{t - 2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d t} (2^{t - 2})

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{t - 2} = e^{t - 2 l n (2)}

= 5 \frac{d}{d t} (e^{t - 2 l n (2)})

Apply the chain rule:

e^{t - 2 l n (2)} \frac{d}{d t} (t - 2 ln (2))

= e^{t - 2 l n (2)} \frac{d}{d t} (t - 2 ln (2))

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

= 5 e^{t - 2 l n (2)} \frac{ln ( 2 )}{2 t - 2}

Simplify

5 e^{t - 2 l n (2)} \frac{ln ( 2 )}{2 t - 2} : \frac{5 ln ( 2 ) \cdot 2 ^{t - 2 - 1}}{t - 2}

= \frac{5 ln ( 2 ) \cdot 2 ^{t - 2 - 1}}{t - 2}

n = 4 \sum \infty e^{7 - 4 n}

\frac{\partial}{\partial x} (sin (x) \cdot cos (x))

x \to + (- 1) lim (x + 2 x^{4})

What is the derivative of 5*2^{sqrt(t-2)} ?
The derivative of 5*2^{sqrt(t-2)} is (5ln(2)*2^{sqrt(t-2)-1})/(sqrt(t-2))
What is the first derivative of 5*2^{sqrt(t-2)} ?
The first derivative of 5*2^{sqrt(t-2)} is (5ln(2)*2^{sqrt(t-2)-1})/(sqrt(t-2))