What is the derivative of y=(sin^3(x)+4)/(3cos^3(x)) ?

The derivative of y=(sin^3(x)+4)/(3cos^3(x)) is (sin(x)(sin^3(x)+cos^2(x)sin(x)+4))/(cos^4(x))

What is the first derivative of y=(sin^3(x)+4)/(3cos^3(x)) ?

The first derivative of y=(sin^3(x)+4)/(3cos^3(x)) is (sin(x)(sin^3(x)+cos^2(x)sin(x)+4))/(cos^4(x))

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derivative of

y = \frac{sin ^{3} ( x ) + 4}{3 cos ^{3} ( x )}

\frac{sin ( x ) ( sin ^{3} ( x ) + cos ^{2} ( x ) sin ( x ) + 4 )}{cos ^{4} ( x )}

Solution steps

\frac{d}{d x} (\frac{sin ^{3} ( x ) + 4}{3 cos ^{3} ( x )})

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (\frac{sin ^{3} ( x ) + 4}{cos ^{3} ( x )})

Apply the Quotient Rule:

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

= \frac{1}{3} \cdot \frac{\frac{d}{d x} ( sin ^{3} ( x ) + 4 ) cos ^{3} ( x ) - \frac{d}{d x} ( cos ^{3} ( x ) ) ( sin ^{3} ( x ) + 4 )}{( cos ^{3} ( x ) ) ^{2}}

\frac{d}{d x} (sin^{3} (x) + 4) = 3 sin^{2} (x) cos (x)

\frac{d}{d x} (cos^{3} (x)) = - 3 cos^{2} (x) sin (x)

= \frac{1}{3} \cdot \frac{3 sin ^{2} ( x ) cos ( x ) cos ^{3} ( x ) - ( - 3 cos ^{2} ( x ) sin ( x ) ) ( sin ^{3} ( x ) + 4 )}{( cos ^{3} ( x ) ) ^{2}}

Simplify

\frac{1}{3} \cdot \frac{3 sin ^{2} ( x ) cos ( x ) cos ^{3} ( x ) - ( - 3 cos ^{2} ( x ) sin ( x ) ) ( sin ^{3} ( x ) + 4 )}{( cos ^{3} ( x ) ) ^{2}} : \frac{sin ( x ) ( sin ^{3} ( x ) + cos ^{2} ( x ) sin ( x ) + 4 )}{cos ^{4} ( x )}

= \frac{sin ( x ) ( sin ^{3} ( x ) + cos ^{2} ( x ) sin ( x ) + 4 )}{cos ^{4} ( x )}

2 x^{3} - 2.5 x - 5

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

cos (6 - 2 x)

t^{2} + 4 t + 5

x^{2} - cos (x) + 2

What is the derivative of y=(sin^3(x)+4)/(3cos^3(x)) ?
The derivative of y=(sin^3(x)+4)/(3cos^3(x)) is (sin(x)(sin^3(x)+cos^2(x)sin(x)+4))/(cos^4(x))
What is the first derivative of y=(sin^3(x)+4)/(3cos^3(x)) ?
The first derivative of y=(sin^3(x)+4)/(3cos^3(x)) is (sin(x)(sin^3(x)+cos^2(x)sin(x)+4))/(cos^4(x))