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

Question

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

Christopher Allers · Accepted Answer

Remember the Quotient Rule by this helpful note:

#dy/dx=(low* d(high) - high* d(low))/ (low^(2))# 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. #dy/dx = ([(x-3)*D_x(3+tanx)]-[(3+tanx)*D_x(x-3)])/(x-3)^(2)# Differentiate all the terms that need to be differentiated. #dy/dx = ([(x-3)*(sec^(2)x)]-[(3+tanx)(1)])/(x-3)^(2)# Further simplifying: #dy/dx= ([(x-3)*(sec^(2)x)]-(3+tanx))/(x-3)^(2)#

Emma Aldridge · Answer

undefined To differentiate $ f(x) = \frac{3 + x \tan(x)}{x - 3} $ using the quotient rule, follow these steps:

1. Identify the numerator and denominator functions: $ u(x) = 3 + x \tan(x) $ and $ v(x) = x - 3 $.
2. Apply the quotient rule formula: $ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - v'(x)u(x)}{(v(x))^2} $.
3. Compute the derivatives of the numerator and denominator functions: $ u'(x) $ and $ v'(x) $.
4. Substitute these derivatives and the original functions into the quotient rule formula.
5. Simplify the expression to obtain the derivative of $ f(x) $.

Applying the quotient rule:

$$ f'(x) = \frac{(3 + x \tan(x)) \cdot (1) - (x - 3) \cdot (1 + \tan^2(x))}{(x - 3)^2} $$

$$ f'(x) = \frac{3 + x \tan(x) - (x - 3)(1 + \tan^2(x))}{(x - 3)^2} $$

$$ f'(x) = \frac{3 + x \tan(x) - (x - 3) - (x - 3)\tan^2(x)}{(x - 3)^2} $$

$$ f'(x) = \frac{3 + x \tan(x) - x + 3 - x\tan^2(x) + 3\tan^2(x)}{(x - 3)^2} $$

$$ f'(x) = \frac{6 + x \tan(x) - x - x\tan^2(x) + 3\tan^2(x)}{(x - 3)^2} $$

$$ f'(x) = \frac{6 - x + x \tan(x) - x\tan^2(x) + 3\tan^2(x)}{(x - 3)^2} $$

$$ f'(x) = \frac{6 - x + x \tan(x) + 2\tan^2(x)}{(x - 3)^2} $$