What is the derivative of (1/2 ^{sqrt(sin^2(x))}) ?

The derivative of (1/2 ^{sqrt(sin^2(x))}) is -(ln(2)sin(2x))/(2^{sin(x)+1)sin(x)}

What is the first derivative of (1/2 ^{sqrt(sin^2(x))}) ?

The first derivative of (1/2 ^{sqrt(sin^2(x))}) is -(ln(2)sin(2x))/(2^{sin(x)+1)sin(x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (\frac{1}{2})^{s i n^{2} (x)}

- \frac{ln ( 2 ) sin ( 2 x )}{2 ^{s i n (x) + 1} sin ( x )}

Solution steps

\frac{d}{d x} (\frac{1}{2})^{s i n^{2} (x)}

Apply exponent rule:

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

(\frac{1}{2})^{s i n^{2} (x)} = e^{s i n^{2} (x) l n (\frac{1}{2})}

= \frac{d}{d x} (e^{s i n^{2} (x) l n (\frac{1}{2})})

Apply the chain rule:

e^{s i n^{2} (x) l n (\frac{1}{2})} \frac{d}{d x} (sin^{2} (x) ln (\frac{1}{2}))

= e^{s i n^{2} (x) l n (\frac{1}{2})} \frac{d}{d x} (sin^{2} (x) ln (\frac{1}{2}))

\frac{d}{d x} (sin^{2} (x) ln (\frac{1}{2})) = - \frac{ln ( 2 ) sin ( 2 x )}{2 sin ^{2} ( x )}

= e^{s i n^{2} (x) l n (\frac{1}{2})} - \frac{ln ( 2 ) sin ( 2 x )}{2 sin ^{2} ( x )}

Simplify

e^{s i n^{2} (x) l n (\frac{1}{2})} - \frac{ln ( 2 ) sin ( 2 x )}{2 sin ^{2} ( x )} : - \frac{ln ( 2 ) sin ( 2 x )}{2 ^{s i n (x) + 1} sin ( x )}

= - \frac{ln ( 2 ) sin ( 2 x )}{2 ^{s i n (x) + 1} sin ( x )}

What is the derivative of (1/2 ^{sqrt(sin^2(x))}) ?
The derivative of (1/2 ^{sqrt(sin^2(x))}) is -(ln(2)sin(2x))/(2^{sin(x)+1)sin(x)}
What is the first derivative of (1/2 ^{sqrt(sin^2(x))}) ?
The first derivative of (1/2 ^{sqrt(sin^2(x))}) is -(ln(2)sin(2x))/(2^{sin(x)+1)sin(x)}