Question

differentiate the following w.r.t. x:

$x\sqrt{a^{2}-x^{2}}+a^{2}sin^{-1}\frac{x}{a}$

Answer 1

Derivative of x √(a² - x²) + a² sin⁻¹(x/a) is {2 √(a² - x²)}.

Step-by-step explanation:

Let, y = x √(a² - x²) + a² sin⁻¹(x/a)

Differentiating both sides with respect to x, we get

dy/dx = d/dx {x √(a² - x²) + a² sin⁻¹(x/a)}

= d/dx {x √(a² - x²)} + a² d/dx {sin⁻¹(x/a)} ..... (1)

Now, d/dx {x √(a² - x²)}

= x d/dx {√(a² - x²)} + √(a² - x²) d/dx (x)

= x d/dx {(a² - x²)^(1/2)} + √(a² - x²)

= x.(1/2).{(a² - x²)^(1/2 - 1)}.d/dx (a² - x²) + √(a² - x²)

= (x/2).{(a² - x²)^(- 1/2)}.(- 2x) + √(a² - x²)

= - x²/√(a² - x²) + √(a² - x²)

= (- x² + a² - x²)/√(a² - x²)

or, d/dx {x √(a² - x²)} = (a² - 2x²)/√(a² - x²)

and d/dx {sin⁻¹(x/a)}

= 1/√{1 - (x/a)²} * d/dx (x/a)

= 1/√(1 - x²/a²) * 1/a

= a/√(a² - x²) * 1/a

or, d/dx {sin⁻¹(x/a)} = 1/√(a² - x²)

From (1), we get

dy/dx = (a² - 2x²)/√(a² - x²) + a²/√(a² - x²)

= (a² - 2x² + a²)/√(a² - x²)

= 2 (a² - x²)/√(a² - x²)

= 2 √(a² - x²),

which is the required derivative.

Answer 2

Proof:

Let y = √(a² - x²) ÷ √(a² + x²) = (a² - x²)½ / (a² + x²)½ ……………………………….(1)

Squaring both sides, y² = (a² - x²)/(a² + x²) ……………………………………………(2)

Taking logarithms on both sides,

log y² = log [(a² - x²)/(a² + x²)]

=[(a² + x²)(-2x) - (a² - x²)(2x)]/(a²+ x²)²

= -2a²x/(a²+ x²)^(2–1/2) . 1/(a² - x²)½ = -2a²x/[(a²+ x²)^3/2 .(a² - x²)½] (Answer)