Derivatives calculus

Question

Find the derivative of the function

 

y=3 tan^-1 (x-(√1+x²))

 

1+x²  is  the square root 

 

Answer 1

Here, d[tan^-1 u]/dx is given as (du/dx)/(1 + u²).

For u = x − (1 + x²)^0.5, u² = x² − 2x(1 + x²)^0.5 + 1 + x² or 2x² − 2x(1 + x²)^0.5 + 1.

Next set down du/dx as 1 − (0.5)(2x)(1 + x²)^-0.5 or 1 − x(1 + x²)^-0.5.

Then (du/dx)/(1 + u²) is assembled as 1 − x(1 + x²)^-0.5 / 1 + 2x² − 2x(1 + x²)^0.5 + 1.

Build the final answer as 3[1 − x(1 + x²)^-0.5] / 2x² − 2x(1 + x²)^0.5 + 2.

Take this last to 1.5[1 − x(1 + x²)^-0.5] / {x² − x(1 + x²)^0.5 + 1} which will simplify to

1.5{(1 + x²)^0.5 − x}

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[(1 + x²)^1.5 − x − x³]