What is the derivative of 2^{sin(x+2022)} ?

The derivative of 2^{sin(x+2022)} is ln(2)*2^{sin(x+2022)}cos(x+2022)

What is the first derivative of 2^{sin(x+2022)} ?

The first derivative of 2^{sin(x+2022)} is ln(2)*2^{sin(x+2022)}cos(x+2022)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

2^{s i n (x + 202 2^{\circ})}

ln (2) \cdot 2^{s i n (x + 202 2^{\circ})} cos (x + 202 2^{\circ})

Solution steps

\frac{d}{d x} (2^{s i n (x + 202 2^{\circ})})

Apply exponent rule:

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

2^{s i n (x + 202 2^{\circ})} = e^{s i n (x + 202 2^{\circ}) l n (2)}

= \frac{d}{d x} (e^{s i n (x + 202 2^{\circ}) l n (2)})

Apply the chain rule:

e^{s i n (x + 202 2^{\circ}) l n (2)} \frac{d}{d x} (sin (x + 202 2^{\circ}) ln (2))

= e^{s i n (x + 202 2^{\circ}) l n (2)} \frac{d}{d x} (sin (x + 202 2^{\circ}) ln (2))

\frac{d}{d x} (sin (x + 202 2^{\circ}) ln (2)) = ln (2) cos (x + 202 2^{\circ})

= e^{s i n (x + 202 2^{\circ}) l n (2)} ln (2) cos (x + 202 2^{\circ})

e^{s i n (x + 202 2^{\circ}) l n (2)} = 2^{s i n (x + 202 2^{\circ})}

= ln (2) \cdot 2^{s i n (x + 202 2^{\circ})} cos (x + 202 2^{\circ})

tan^{4} (5 x)

\frac{1}{( x ) ^{\frac{1}{2}}}

\frac{2}{( x - 3 )}

y = sin^{\frac{1}{2}} (2 x + 1)

2^{x} ln^{2} (x)

What is the derivative of 2^{sin(x+2022)} ?
The derivative of 2^{sin(x+2022)} is ln(2)*2^{sin(x+2022)}cos(x+2022)
What is the first derivative of 2^{sin(x+2022)} ?
The first derivative of 2^{sin(x+2022)} is ln(2)*2^{sin(x+2022)}cos(x+2022)