If y - -x2 - 1x - 1 - then find dydx-

Here, u(x) = x^2 + 1, v(x) = x + 1 .Using quotient rule, we get dydx = (x + 1)d(x^2 +1)dx (x^2 + 1)d(x + 1)dx(x+ 1)^2 = (x + 1) 2x (x^2 + 1)1( ...

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Exponents with Like Bases

If y = [x2 ...

Question

If $y = [\frac{x^{2} + 1}{x + 1}]$ , then find $\frac{d y}{d x}$ .

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Solution

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y = x log ∣ ∣ x + \sqrt{(1 + x^{2})} ∣ ∣ - \sqrt{1 + x^{2}}

\frac{d y}{d x} =

y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . . . . . . \infty

\frac{d y}{d x}

y = {sin}^{- 1} [x \sqrt{x - 1} - \sqrt{x} \sqrt{1 - x^{2}}]

\frac{d y}{d x}

y = {tan}^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}

\frac{d y}{d x} =

y \sqrt{1 - x^{2}} + x \sqrt{1 - y^{2}} = 1

\frac{d y}{d x}

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