Expert: Scotto - 6/17/2010

Question
Hi, I was wondering if maybe you could help me with 2 questions in regards to derivatives.

a) g(x) = (sqrtx^2-1)*sec^-1x

b) y = tan^-1(x-sqrt1+x^2)

best regards,

Umair

Answer
a) This is a product rule between f(x) and h(x) where f(x) = (x²-1)^(1/2) and h(x) = sec^-1(x).

Think of the function sqrt(x) = x^(1/2).
The derivative is then (1/2)x^(-1/2).

Applying this to f(x) involves a chain rule, so if f(x) = (x²-1)^(1/2),
f'(x) = (1/2)[(x²-1)^(-1/2)][d(x²-1)/dx].
It can be seen that d(x²-1)/dx = 2x, so f'(x) = (1/2)[(x²-1)^(-1/2)]2x.
The negative power puts the number in the denominator with a positive exponet and 1/2 * 2 is 1,
so we have f'(x) = x/(x²-1)^0.5.

Thus, we have f(x) = (x²-1)^(1/2) and f'(x) = x/(x²-1)^0.5.

Now as far as finding h'(x), instead of using y = sec^-1(x), lets say sec(y) = x.
Taking the derivative gives sec(y)tan(y) dy/dx = 1, do dy/dx = 1/(sec(y)tan(y)).

If sec(y) = x, then we can take the hypotenuse as x, the near side as 1, and the farside becomes
√(x²-1) since 1² = 1.  Note that the angle in the triangle is y.  This makes tan(y) = √(x²-1)/1,
so we get tan(y)= √(x²-1).

We can now take dy/dx = 1/sec(y)tan(y) and say that it is really dy/dx = 1/[x√(x²-1)].

Thus, we can say g(x) = sec^-1(x) and g'(x) = 1/[x√(x²-1)].

We can now find g'(x) as g'(x) = f(x)g'(x) + f'(x)g(x).


b) Let tan(y) = x - √(1+x²).
Taking the derivative of this equation with respect to x gives
sec²(y)(dy/dx) = 1 - d[√(1+x²)]/dx.
Note that to compute d[√(1+x²)]/dx involves the chain rule of d[(1+x²)^(1/2)]/dx.
It can be seen, using this, that d[(1+x²)^(1/2)]/dx = (1/2)(1+x²)^(-1/2)(2x) = x/√(1+x²).

That makes the equation sec²(y)(dy/dx) = 1 - d[√(1+x²)]/dx into sec²(y)(dy/dx) = 1 - x/√(1+x²).

Take a triangle and let the far side be x - √(1+x²),
so the near side is 1 and the angle is y.
This makes the hypotenuse into √(1² + [x - √(1+x²)]²).
Now [x - √(1+x²)]² = x² - 2x√(1+x²) + 1+x² = 2x² - 2x√(1+x²) + 1,
so adding this to 1 in the squareroot gives √(2x² - 2x√(1+x²) + 2).

Now sec(y) = near side / hypotenuse, so sec(y) = 1/√(2x² - 2x√(1+x²) + 2).
This makes sec²(y) = 1/(2x² - 2x√(1+x²) + 2).
If we divide both sides of sec²(y)(dy/dx) = 1 - x/√(1+x²) by sec²(y), this gives us
dy/dx = [1 - x/√(1+x²)]/{1/(2x² - 2x√(1+x²) + 2)}.
Invert and multiply the bottom 1/(2x² - 2x√(1+x²) + 2) to get
dy/dx = [1 - x/√(1+x²)](2x² - 2x√(1+x²) + 2).

One final thing is a 2 can be factored out of (2x² - 2x√(1+x²) + 2), giving
dy/dx =  2[1 - x/√(1+x²)](x² - x√(1+x²) + 1).

Now weird math problems like that can get confusing.  Does that look right to you?
I sometimes slip up on my arithmetic, but that is the conept.