What is the derivative of (sin(t)/(sqrt(1+x^2))) ?

The derivative of (sin(t)/(sqrt(1+x^2))) is -(xsin(t))/((1+x^2)^{3/2)}

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

The first derivative of (sin(t)/(sqrt(1+x^2))) is -(xsin(t))/((1+x^2)^{3/2)}

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\frac{d}{d x} (\frac{sin ( t )}{1 + x ^{2}})

- \frac{x sin ( t )}{( 1 + x ^{2} ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{sin ( t )}{1 + x ^{2}})

Treat

t

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

- \frac{1}{2 ( 1 + x ^{2} ) ^{\frac{3}{2}}} \frac{d}{d x} (1 + x^{2})

= - \frac{1}{2 ( 1 + x ^{2} ) ^{\frac{3}{2}}} \frac{d}{d x} (1 + x^{2})

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

= sin (t) (- \frac{1}{2 ( 1 + x ^{2} ) ^{\frac{3}{2}}} \cdot 2 x)

Simplify

sin (t) (- \frac{1}{2 ( 1 + x ^{2} ) ^{\frac{3}{2}}} \cdot 2 x) : - \frac{x sin ( t )}{( 1 + x ^{2} ) ^{\frac{3}{2}}}

= - \frac{x sin ( t )}{( 1 + x ^{2} ) ^{\frac{3}{2}}}

\int x^{3} e^{a x^{4}} d x

\int \frac{2 ( x ^{2} + 1 )}{x ( x ^{2} + 2 )} d x

x \to 4 - lim (\frac{1}{x ^{2} - 16})

\int \frac{x}{( 1 - x ^{2} )} d x

y = ln (x^{2} - 4 x + 1), (4, 0)

What is the derivative of (sin(t)/(sqrt(1+x^2))) ?
The derivative of (sin(t)/(sqrt(1+x^2))) is -(xsin(t))/((1+x^2)^{3/2)}
What is the first derivative of (sin(t)/(sqrt(1+x^2))) ?
The first derivative of (sin(t)/(sqrt(1+x^2))) is -(xsin(t))/((1+x^2)^{3/2)}