Find the derivative of the function below.

Question

Find the derivative of the function below.

A) f(x)= (sqrt x^2-9 sec^-1 (x))

Answer 1

f(x) = √(x²-9) sec^-1x

Using the product rule, f'(x) = (1/2)(x²-9)^-1/2(2x)sec^-1x + [1 / (lxl√(x²-1))]√(x²-9)

= [x / √(x²-9)]sec^-1x + √(x²-9) / (lxl√(x²-1))

Answer 2

f(x) = √(x²–9) × sec⁻¹(x) = u × v where u = √(x²–9) and v = sec⁻¹(x)

To find the derivative we use the product rule

f’(x) = d(u×v)dx = v×(du/dx) + u×(dv/dx)

du/dx = [1/(2√(x²–9))] × 2x = 2x/2√(x²–9) = x/√(x²–9)

so v ×(du/dx) = x sec⁻¹(x)/√(x²–9)

dv/dx = 1/|x|√(x²–1)

so u×(dv/dx) = √(x²–9)/|x|√(x²–1)

and

v×(du/dx) + u×(dv/dx)= x sec⁻¹(x)/√(x²–9) + √(x²–9)/|x|√(x²–1)

Answer

f’(x) = x sec⁻¹(x)/√(x²–9) + √(x²–9)/|x|√(x²–1)

Note

The derivative of sec⁻¹(x) comes from:

First noting that y = sec⁻¹(x) means that sec y = x

Then differentiating both sides to give

sec y tan y (dy/dx) = dxdx = 1, so dy/dx = 1/(sec y tan y)

And finally substituting back to write dy/dx as a function of x

sec y = x and tan y = √( sec²y – 1) = √( x² – 1)

This gives the derivative of sec⁻¹(x) = dy/dx = 1/x√(x² – 1)

But this is not entirely correct.

If you look at plot of y = sec⁻¹(x) (where the domain is |x| ≥ 1) you see that the slope of the curve is always positive. So the formula above is in need of a correction:

The x in the denomination need to be replaced by |x|.

So the derivative of sec⁻¹(x) = 1/|x|√(x² – 1)