What is the derivative of 3cos(x+4x^2) ?

The derivative of 3cos(x+4x^2) is -3sin(x)+8x

What is the first derivative of 3cos(x+4x^2) ?

The first derivative of 3cos(x+4x^2) is -3sin(x)+8x

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

- 3 sin (x) + 8 x

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (4 x^{2}) = 8 x

= - 3 sin (x) + 8 x

x^{4} + 5 y^{4} x^{5} + 4 y^{^{'}} = 0, y (0) = 1

\int \frac{4 x ^{4} - 5 x ^{2}}{( x + 3 ) ( x - 2 )} d x

\int x sin (n) x d x

n \to 2 lim (\frac{n - 1}{n ^{2} - 1})

\frac{d}{d x} (x^{3} - 3 x^{2} - 9 x + 5)

What is the derivative of 3cos(x+4x^2) ?
The derivative of 3cos(x+4x^2) is -3sin(x)+8x
What is the first derivative of 3cos(x+4x^2) ?
The first derivative of 3cos(x+4x^2) is -3sin(x)+8x