Question

Find the derivative using quiotent rule ​

Answer 1

`\frac{\text{d}y}{\text{d}x}=(1)/(2√(x)(√(x)+1)^2)`

Explanation:

Given rational equation:

`y=(√(x))/(√(x)+1)`

To find the derivative of the given equation, we can use the quotient rule.

`\boxed{\begin{array}{c}\underline{\textsf{Quotient Rule for Differentiation}}\\\\\textsf{If $y=(u)/(v)$ then:}\\\\\frac{\text{d}y}{\text{d}x}=\frac{v \frac{\text{d}u}{\text{d}x}-u\frac{\text{d}v}{\text{d}x}}{v^2}\\\\\end{array}}`

First, identify u and v and differentiate them separately:

`\textsf{Let}\;\;u=√(x)=x^{(1)/(2)}\implies \frac{\text{d}u}{\text{d}x}=(1)/(2)x^{-(1)/(2)}=(1)/(2√(x))`

`\textsf{Let}\;\;v=√(x)+1=x^{(1)/(2)}+1 \implies \frac{\text{d}v}{\text{d}x}=(1)/(2)x^{-(1)/(2)}+0=(1)/(2√(x))`

Now, put everything into the quotient rule formula:

`\frac{\text{d}y}{\text{d}x}=((√(x)+1) \cdot (1)/(2√(x))-√(x) \cdot (1)/(2√(x)))/((√(x)+1)^2)`

Simplify where possible:

`\frac{\text{d}y}{\text{d}x}=((√(x)+1)/(2√(x))-(√(x))/(2√(x)))/((√(x)+1)^2)`

`\frac{\text{d}y}{\text{d}x}=((√(x)+1-√(x))/(2√(x)))/((√(x)+1)^2)`

`\frac{\text{d}y}{\text{d}x}=((1)/(2√(x)))/((√(x)+1)^2)`

`\frac{\text{d}y}{\text{d}x}=(1)/(2√(x)(√(x)+1)^2)`

Therefore, the derivative of the given equation is:

`\large\boxed{\boxed{\frac{\text{d}y}{\text{d}x}=(1)/(2√(x)(√(x)+1)^2)}}`