Question

assume that ( f(x) ) is differentiable. find an expression for the derivative of ( y ) at ( x = 4 ), assuming that ( f(4)=-3 ) and ( f^{prime}(4)=3 )
( y=\frac{f(x)+1}{f(x)+x} )
the derivative of ( y ) at ( x = 4 ) is ( square ).

Explanation:

Step1: Apply quotient rule

The quotient rule states that if \(y = \frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here, \(u = f(x)+1\), so \(u^\prime=f^\prime(x)\); \(v = f(x)+x\), so \(v^\prime=f^\prime(x)+1\). Then \(y^\prime=\frac{f^\prime(x)(f(x)+x)-(f(x)+1)(f^\prime(x)+1)}{(f(x)+x)^{2}}\).

Step2: Substitute \(x = 4\)

Given \(f(4)=- 3\) and \(f^\prime(4)=3\).
First, find \(u(4)=f(4)+1=-3 + 1=-2\), \(v(4)=f(4)+4=-3 + 4 = 1\), \(u^\prime(4)=f^\prime(4)=3\), \(v^\prime(4)=f^\prime(4)+1=3 + 1=4\).
Then \(y^\prime(4)=\frac{3\times1-(-2)\times4}{1^{2}}\).
Calculate the numerator: \(3\times1-(-2)\times4=3 + 8=11\).

Answer:

\(11\)