Question

find derivative of y= 1/(3x^2+2x+1)

Answer 1

Explanation:

y= 1/(3x^2+2x+1)

y=(3x²+2x+1)^-1

dy/dx=-1(3x²+2x+1)^-2×d/dx(3x²+2x+1)

dy/dx=-1(3x²+2x+1)^-2×(6x+2)

dy/dx=-(6x+2)/(3x²+2x+1)²

Answer 2

Answer:

Required derivative is $- \frac{2(3x + 1)}{ {(3 {x}^{2} + 2x + 1)}^{2} }$

Explanation:

Given,

$y = \frac{1}{3 {x}^{2} + 2x + 1}$

Here we want to find derivative of y with respect to x.

We are differentiating y with respect to x,

$\frac{dy}{dx} = \frac{d}{dx} ( \frac{1}{3 {x}^{2} + 2x + 1} ) \\ = - { \frac{1}{(3 {x}^{2} + 2x + 1 )^{2}} } \times \frac{d}{dx} (3 {x}^{2} + 2x + 1)$

$= - { \frac{1}{(3 {x}^{2} + 2x + 1 )^{2}} } \times (3 \times 2 {x} + 2 ) \\ = - { \frac{1}{(3 {x}^{2} + 2x + 1 )^{2}} } \times (6 {x} + 2) \\ = - \frac{2(3x + 1)}{ {(3 {x}^{2} + 2x + 1)}^{2} }$

Here applied formula,

$\frac{d}{dx} ( {x}^{a} ) = a {x}^{a - 1}$

Some more important Derivatives formula,

$1. \frac{d}{dx} ( \sin(x) ) = \cos(x ) \\ 2. \frac{d}{dx} ( \cos(x) ) = - \sin(x) \\ 3. \frac{d}{dx} ( \tan(x) ) = {sec}^{2} x \\ 4. \frac{d}{dx} ( \cot(x) ) = - {cosec}^{2} x \\ 5. \frac{d}{dx} ( \sec(x) ) = \sec(x) \tan(x) \\ 6. \frac{d}{dx} (cosec(x)) = - cosecxcotx$

Know more about Derivative

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