How do you find the derivatives of $y=((x^2+1)sqrt(3x+4))/((2x-3)sqrt(x^2-4))$ by logarithmic differentiation?

Question

6489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T01:28:02

Take the natural logarithm of both sides.

$lny= ln(((x^2 + 1)(sqrt(3x + 4)))/((2x - 3)sqrt(x^2 - 4)))$

Use the rules $ln(ab) = lna + lnb$ and $ln(a/b) = lna - lnb$ :

$lny = ln((x^2 + 1)sqrt(3x + 4)) - ln((2x - 3)sqrt(x^2 - 4))$

$lny = ln(x^2 + 1) + ln(3x + 4)^(1/2) - (ln(2x - 3) + ln(x^2 - 4)^(1/2))$

Now use the rule $lnn^a = alnn$ to simplify further.

$lny = ln(x^2 + 1) + 1/2ln(3x + 4) - ln(2x - 3) - 1/2ln(x^2 - 4)$

Now differentiate using the chain rule.

$1/y(dy/dx) = (2x)/(x^2 + 1) + 3/(2(3x + 4)) - 2/(2x - 3) - (2x)/(2(x^2 - 4))$

$1/y(dy/dx) = (2x)/(x^2 + 1) + 3/(2(3x + 4)) - 2/(2x - 3) - x/(x^2 - 4)$

$dy/dx = y((2x)/(x^2 + 1) + 3/(2(3x +4)) - 2/(2x - 3) - x/(x^2 - 4))$

$dy/dx= ((x^2 + 1)sqrt(3x + 4))/((2x - 3)(sqrt(x^2 - 4)))((2x)/(x^2 + 1) + 3/(2(3x + 4)) - 2/(2x- 3) - x/(x^2 - 4)$

This can probably be simplified further, but I'll leave those workings up to you:).

Hopefully this helps!

How do you find the derivatives of y=((x^2+1)sqrt(3x+4))/((2x-3)sqrt(x^2-4))y=((x^2+1)sqrt(3x+4))/((2x-3)sqrt(x^2-4)) by logarithmic differentiation?