Question

use the formula ( f^{prime}(x)=lim _{z
ightarrow x} \frac{f(z)-f(x)}{z - x} ) to find the derivative of ( f(x)=\frac{7 x}{x + 1} ).
the derivative of ( f(x)=\frac{7 x}{x + 1} ) is ( square ).

Explanation:

Step1: Substitute \(f(z)\) and \(f(x)\) into the formula

$$
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$$

Step2: Simplify the numerator

$$
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$$

So, \(f^{\prime}(x)=\lim_{z
ightarrow x}\frac{7(z - x)}{(z + 1)(x + 1)(z - x)}\)

Step3: Cancel out \((z - x)\)

Since \(z
eq x\) (as we are taking the limit \(z
ightarrow x\) but \(z
eq x\) in the expression \(\frac{7(z - x)}{(z + 1)(x + 1)(z - x)}\)), we can cancel out \((z - x)\)
\(f^{\prime}(x)=\lim_{z
ightarrow x}\frac{7}{(z + 1)(x + 1)}\)

Step4: Evaluate the limit

Substitute \(z = x\) into \(\frac{7}{(z + 1)(x + 1)}\)
\(f^{\prime}(x)=\frac{7}{(x + 1)(x + 1)}=\frac{7}{(x + 1)^{2}}\)

Answer:

\(\frac{7}{(x + 1)^{2}}\)