What is the d/(da)(((sin^2(a)))/((1+cos^2(a)))) ?

The d/(da)(((sin^2(a)))/((1+cos^2(a)))) is (2sin(2a))/((1+cos^2(a))^2)

What is the first d/(da)(((sin^2(a)))/((1+cos^2(a)))) ?

The first d/(da)(((sin^2(a)))/((1+cos^2(a)))) is (2sin(2a))/((1+cos^2(a))^2)

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\frac{d}{d a} (\frac{( sin ^{2} ( a ) )}{( 1 + cos ^{2} ( a ) )})

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

Solution steps

\frac{d}{d a} (\frac{( sin ^{2} ( a ) )}{( 1 + cos ^{2} ( a ) )})

Apply the Quotient Rule:

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

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

\frac{d}{d a} (sin^{2} (a)) = sin (2 a)

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

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

Simplify

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

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

\frac{e ^{1}}{x}

(201 6^{u}) (u + 2017)

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

\frac{d}{d t} (\frac{( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t )})

e^{3 x + 2} + 5

What is the d/(da)(((sin^2(a)))/((1+cos^2(a)))) ?
The d/(da)(((sin^2(a)))/((1+cos^2(a)))) is (2sin(2a))/((1+cos^2(a))^2)
What is the first d/(da)(((sin^2(a)))/((1+cos^2(a)))) ?
The first d/(da)(((sin^2(a)))/((1+cos^2(a)))) is (2sin(2a))/((1+cos^2(a))^2)