What is the derivative of f(x)=(cos(e^x))/(x^4+1) ?

The derivative of f(x)=(cos(e^x))/(x^4+1) is (-e^xsin(e^x)(x^4+1)-4x^3cos(e^x))/((x^4+1)^2)

What is the first derivative of f(x)=(cos(e^x))/(x^4+1) ?

The first derivative of f(x)=(cos(e^x))/(x^4+1) is (-e^xsin(e^x)(x^4+1)-4x^3cos(e^x))/((x^4+1)^2)

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derivative of

f (x) = \frac{cos ( e ^{x} )}{x ^{4} + 1}

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

Solution steps

\frac{d}{d x} (\frac{cos ( e ^{x} )}{x ^{4} + 1})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

2 x, \frac{2 x}{3}

\int \frac{x ^{2} + 1}{x 1 + x ^{4}} d x

\frac{d}{d t} (e^{2 t} cos (2 t))

8 x^{4} ln (6 x)

\int cot (x) cos^{2} (x) d x

What is the derivative of f(x)=(cos(e^x))/(x^4+1) ?
The derivative of f(x)=(cos(e^x))/(x^4+1) is (-e^xsin(e^x)(x^4+1)-4x^3cos(e^x))/((x^4+1)^2)
What is the first derivative of f(x)=(cos(e^x))/(x^4+1) ?
The first derivative of f(x)=(cos(e^x))/(x^4+1) is (-e^xsin(e^x)(x^4+1)-4x^3cos(e^x))/((x^4+1)^2)