What is the derivative of f(x)=sqrt(3x^2+3x+4) ?

The derivative of f(x)=sqrt(3x^2+3x+4) is (6x+3)/(2sqrt(3x^2+3x+4))

What is the first derivative of f(x)=sqrt(3x^2+3x+4) ?

The first derivative of f(x)=sqrt(3x^2+3x+4) is (6x+3)/(2sqrt(3x^2+3x+4))

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derivative of

f (x) = 3 x^{2} + 3 x + 4

\frac{6 x + 3}{2 3 x ^{2} + 3 x + 4}

Solution steps

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

Apply the chain rule:

\frac{1}{2 3 x ^{2} + 3 x + 4} \frac{d}{d x} (3 x^{2} + 3 x + 4)

= \frac{1}{2 3 x ^{2} + 3 x + 4} \frac{d}{d x} (3 x^{2} + 3 x + 4)

\frac{d}{d x} (3 x^{2} + 3 x + 4) = 6 x + 3

= \frac{1}{2 3 x ^{2} + 3 x + 4} (6 x + 3)

Simplify

\frac{1}{2 3 x ^{2} + 3 x + 4} (6 x + 3) : \frac{6 x + 3}{2 3 x ^{2} + 3 x + 4}

= \frac{6 x + 3}{2 3 x ^{2} + 3 x + 4}

- x^{2} + 6 x

\frac{d}{d x} (\frac{8}{1 + x})

f (x) = ln (tan (x))

\frac{d x}{d t} = \frac{6 x ^{2} + t t ^{2} + x ^{2}}{6 t x}

\int_{0}^{3} (- x^{3} + 3 x^{2} - 2) d x

What is the derivative of f(x)=sqrt(3x^2+3x+4) ?
The derivative of f(x)=sqrt(3x^2+3x+4) is (6x+3)/(2sqrt(3x^2+3x+4))
What is the first derivative of f(x)=sqrt(3x^2+3x+4) ?
The first derivative of f(x)=sqrt(3x^2+3x+4) is (6x+3)/(2sqrt(3x^2+3x+4))