Differentiate root x^2+1/(x-9)^2?

Question

With full working out plz ASAP thanks

Answer 1

= √(x² + 1) / (x - 9)²

Quotient rule: (d/dx)[g(x)/h(x)] =[ h(x) g'(x) - g(x) h'(x)] / h(x)²

={ (x - 9)² √(x² + 1)' - √(x² + 1)[(x - 9)²]'} / [(x - 9)²]²

General derivative formulas: (d/dx)(√x) = 1/(2√x) and (d/dx)(aⁿ) = na^n-1

= { (x - 9)² [1/2√(x² + 1)](2x) - √(x² + 1)[2(x - 9)](1)} / [(x - 9)²]²

= { (x - 9)² [1/√(x² + 1)](x) - √(x² + 1)[2(x - 9)]} / [(x - 9)²]²

= {[ x(x - 9)² / √(x² + 1)] - 2(x - 9)√(x² + 1) } / [(x - 9)²]²

Take out common factors.

= (x-9){[ x(x - 9) / √(x² + 1)] - 2√(x² + 1)} / (x - 9)⁴

= {[ x(x - 9) / √(x² + 1)] - 2√(x² + 1)} / (x - 9)³

Rewrite the expression with common denominator.

= 1/√(x² + 1){[ x(x - 9) - 2 √(x² + 1)√(x² + 1)} / (x - 9)³

= 1/√(x² + 1){( x² - 9x) - 2 (x² + 1)} / (x - 9)³

={( x² - 9x) - 2 (x² + 1)} / √(x² + 1)(x - 9)³

= (x² - 9x - 2x² - 2) / [√(x² + 1)(x - 9)³]

= (- x² - 9x - 2) / [(x - 9)³√(x² + 1)]