How do you differentiate f(x)= ( 3+ x tanx )/ (x -3)  using the quotient rule?

Sep 26, 2016

Remember the Quotient Rule by this helpful note:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{l o w \cdot d \left(h i g h\right) - h i g h \cdot d \left(l o w\right)}{l o {w}^{2}}$

Explanation:

For the denominator:
1) Square the function in the denominator

For the numerator:
1) LOW - Copy the "low function" ( the function in the denominator)
2) D(HIGH)- Multiply this (1) by the "d(high)" (the derivative of the function in the numerator)
3) Put minus. This is division so we have to subtract.
4) HIGH - Copy the "high function" (the function in the numerator)
5) D(LOW) - Multiply this (4) by the "d(low)" (the derivative of the function in the denominator

By this, we can now find the derivative using the Quotient Rule.

Setup the derivative. The denominator will be automatically squared.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\left(x - 3\right) \cdot {D}_{x} \left(3 + \tan x\right)\right] - \left[\left(3 + \tan x\right) \cdot {D}_{x} \left(x - 3\right)\right]}{x - 3} ^ \left(2\right)$

Differentiate all the terms that need to be differentiated.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\left(x - 3\right) \cdot \left({\sec}^{2} x\right)\right] - \left[\left(3 + \tan x\right) \left(1\right)\right]}{x - 3} ^ \left(2\right)$

Further simplifying:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\left(x - 3\right) \cdot \left({\sec}^{2} x\right)\right] - \left(3 + \tan x\right)}{x - 3} ^ \left(2\right)$