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Derivative of −2xx2+1- \frac{2 x}{x^{2} + 1}

The calculator will find the derivative of −2xx2+1- \frac{2 x}{x^{2} + 1}, with steps shown.

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Your Input

Find ddx(−2xx2+1)\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right).

Solution

Apply the constant multiple rule ddx(cf(x))=cddx(f(x))\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right) with c=−2c = -2 and f(x)=xx2+1f{\left(x \right)} = \frac{x}{x^{2} + 1}:

(ddx(−2xx2+1))=(−2ddx(xx2+1)){\color{red}\left(\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right)\right)} = {\color{red}\left(- 2 \frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}

Apply the quotient rule ddx(f(x)g(x))=ddx(f(x))g(x)−f(x)ddx(g(x))g2(x)\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}} with f(x)=xf{\left(x \right)} = x and g(x)=x2+1g{\left(x \right)} = x^{2} + 1:

−2(ddx(xx2+1))=−2(ddx(x)(x2+1)−xddx(x2+1)(x2+1)2)- 2 {\color{red}\left(\frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)} = - 2 {\color{red}\left(\frac{\frac{d}{dx} \left(x\right) \left(x^{2} + 1\right) - x \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}

The derivative of a sum/difference is the sum/difference of derivatives:

−2(−x(ddx(x2+1))+(x2+1)ddx(x))(x2+1)2=−2(−x(ddx(x2)+ddx(1))+(x2+1)ddx(x))(x2+1)2- \frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(- x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)} + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2}}

Apply the power rule ddx(xn)=nxn−1\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} with n=2n = 2:

−2(−x((ddx(x2))+ddx(1))+(x2+1)ddx(x))(x2+1)2=−2(−x((2x)+ddx(1))+(x2+1)ddx(x))(x2+1)2- \frac{2 \left(- x \left({\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(1\right)\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(- x \left({\color{red}\left(2 x\right)} + \frac{d}{dx} \left(1\right)\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2}}

The derivative of a constant is 00:

−2(−x(2x+(ddx(1)))+(x2+1)ddx(x))(x2+1)2=−2(−x(2x+(0))+(x2+1)ddx(x))(x2+1)2- \frac{2 \left(- x \left(2 x + {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(- x \left(2 x + {\color{red}\left(0\right)}\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x\right)\right)}{\left(x^{2} + 1\right)^{2}}

Apply the power rule ddx(xn)=nxn−1\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} with n=1n = 1, in other words, ddx(x)=1\frac{d}{dx} \left(x\right) = 1:

−2(−2x2+(x2+1)(ddx(x)))(x2+1)2=−2(−2x2+(x2+1)(1))(x2+1)2- \frac{2 \left(- 2 x^{2} + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(- 2 x^{2} + \left(x^{2} + 1\right) {\color{red}\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2}}

Simplify:

−2(1−x2)(x2+1)2=2(x2−1)(x2+1)2- \frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right)^{2}} = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}

Thus, ddx(−2xx2+1)=2(x2−1)(x2+1)2\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right) = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}.

Answer

ddx(−2xx2+1)=2(x2−1)(x2+1)2\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right) = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}A

Related Notes: Definition of Derivative , Derivatives of Elementary Functions , Table of Derivatives , Related Rates , Studying Derivative Graphically , Optimization Problems , Applications to Economics , Constant Multiple Rule , Sum and Difference Rules , Product Rule , Quotient Rule , Chain Rule , Derivative of Inverse Function