Question

Derivative of these questions
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Answer 1

Answer:
`\bold{y'=(1)/(2(a-b)^2)\bigg((1)/(√(x+a))+(1)/(√(x+b))\bigg)}`

Explanation:

`y=(1)/(√(x+a)-√(x+b))\\\\\\\text{Rationalize the denominator:}\\y=(1)/(√(x+a)-√(x+b))\bigg((√(x+a)+√(x+b))/(√(x+a)+√(x+b))\bigg)\\\\\\y=(√(x+a)+√(x+b))/(x+a-x-b)\\\\\\y=(√(x+a)+√(x+b))/(a-b)\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}`

`\text{Derivative of }√(x+a)\quad \rightarrow \quad (x+a)^{(1)/(2)}\quad = \quad (1)/(2)(x + a)^{-(1)/(2)}\\\\\text{Derivative of }√(x+b)\quad \rightarrow \quad (x+b)^{(1)/(2)}\quad = \quad (1)/(2)(x + b)^{-(1)/(2)}\\\\\\\text{Derivative of}\ (√(x+a)+√(x+b))/(a-b):\\\\= \frac{(1)/(2)(x + a)^{-(1)/(2)}+(1)/(2)(x + b)^{-(1)/(2)}}{(a-b)^2}`

`\text{Factor out }(1)/(2)\ \text{from the numerator:}\\\\\frac{(x+a)^{-(1)/(2)}+(x+b)^{-(1)/(2)}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=(1)/(2(a-b)^2)\bigg((1)/(√(x+a))+(1)/(√(x+b))\bigg)`