Question

If anyone could help me, I'll really appreciate it.

Differentiate the following functions with respect to x.

$y = {cosh}^{ - 1} (2x + 1) - {xsech}^{ - 1} (x)$
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Answer 1

Answer:

$\frac{d y}{d x} = \frac{2}{\sqrt{(2 x+1)^{2} -1} } + (\frac{-x}{|x|\sqrt{1-x^2}) }) + (1) Sec h^{-1} (x)$

Step-by-step explanation:

Step(i):-

Given function

               $y = cosh^{-1} (2 x +1) - x Sec h^{-1} (x)$   ....(i)

we will use differentiation formulas

i) y = cos h⁻¹ (x)

   Derivative of cos h⁻¹ (x)

           $\frac{d y}{d x} = \frac{1}{\sqrt{x^2-1} }$

ii)

      y = sec h⁻¹ (x)

   Derivative of sec h⁻¹ (x)

           $\frac{d y}{d x} = \frac{-1}{|x|\sqrt{(x^2-1} }$

Apply U V formula

$\frac{d UV}{d x} = U V^{l} + V U^{l}$

Step(ii):-

Differentiating equation (i) with respective to 'x'

$\frac{d y}{d x} = \frac{1}{\sqrt{(2 x+1)^{2} -1} } X \frac{d}{d x} (2 x+1) + x (\frac{-1}{|x|\sqrt{1-x^2}) }) + (1) Sec h^{-1} (x)$

$\frac{d y}{d x} = \frac{1}{\sqrt{(2 x+1)^{2} -1} } X (2) + (\frac{-x}{|x|\sqrt{1-x^2}) }) + (1) Sec h^{-1} (x)$

Conclusion:-

$\frac{d y}{d x} = \frac{2}{\sqrt{(2 x+1)^{2} -1} } + (\frac{-x}{|x|\sqrt{1-x^2}) }) + (1) Sec h^{-1} (x)$