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Popular >

((sin(x))/(1+x))^'

Solution

(\frac{sin ( x )}{1 + x})^{^{'}}

Solution

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

Solution steps

(\frac{sin ( x )}{1 + x})^{'}

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

(sin (x))^{'} = cos (x)

(1 + x)^{'} = 1

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

Simplify

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

Graph

Popular Examples

y^{''}+2y^'+2y=0,y(0)=-2,y^'(0)=-4

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = - 2, y^{^{'}} (0) = - 4

(-x+2)*y^'-y=e^{-2x}

(- x + 2) \cdot y^{^{'}} - y = e^{- 2 x}

(dy)/(dx)=a(y^2+1)

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

derivative of (x^2/(1-x^2))

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

(dy)/(dx)=(1+x)(1+y^2)

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

Frequently Asked Questions (FAQ)

What is the ((sin(x))/(1+x))^' ?
The ((sin(x))/(1+x))^' is (cos(x)(1+x)-sin(x))/((1+x)^2)
What is the first ((sin(x))/(1+x))^' ?
The first ((sin(x))/(1+x))^' is (cos(x)(1+x)-sin(x))/((1+x)^2)