Question

Determine the derivative with quotient rule and find the equation of the tangent to f(x) at x=0

Answer 1

`\text{ The derivative of the function is; }\frac{\text{ dy}}{\text{ dx}}\text{ = }\frac{\text{ -2x}^2\text{ + 2}}{\text{ \lparen x}^2\text{ + 1\rparen}^2}`

Step-by-step explanation

Given that;

`\text{ f\lparen x\rparen = }\frac{\text{ 2x}}{\text{ x}^2\text{ + 1}}`

Let U(x) = 2x and V(x) = x^2 + 1

Apply the quotient rule to find the derivative of f(x)

`\text{ }\frac{\text{ dy}}{\text{ dx}}\text{ = }\frac{\text{ V}\frac{\text{ dU}}{\text{ dx}}\text{ - U }\frac{dv}{\text{ dx}}}{\text{ V}^2}`

Differentiate U and V with respect to x

`\begin{gathered} \text{ U\lparen x\rparen = 2x} \\ \text{ }\frac{\text{ du}}{\text{ dx}}\text{ = 2} \\ \\ \\ \text{ V\lparen x\rparen= x}^2\text{ + 1} \\ \text{ }\frac{dv}{\text{ dx}}\text{ = 2x} \end{gathered}`

Hence, we have

`\begin{gathered} \text{ }\frac{\text{ dy}}{\text{ dx}}\text{ = }\frac{(x^2\text{ + 1\rparen}*2\text{ - 2x\lparen2x\rparen}}{(x^2\text{ + 1\rparen}^2} \\ \\ \text{ }\frac{\text{ dy}}{\text{ dx}}\text{ = }\frac{\text{ 2x}^2\text{ + 2 - 4x}^2}{(x^2\text{ + 1\rparen}^2} \\ \\ \text{ }\frac{\text{ dy}}{\text{ dx}}\text{ = }\frac{\text{ -2x}^2\text{ + 2}}{\text{ \lparen x}^2\text{ + 1\rparen}^2} \end{gathered}`