use the formula ( f^{prime}(x)=lim _{z \rightarrow x} \frac{f(z)-f(x)}{z - x} ) to find the derivative of (…

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

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

Answer

Explanation:

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

[ \begin{align*} f^{\prime}(x)&=\lim_{z\rightarrow x}\frac{\frac{7z}{z + 1}-\frac{7x}{x + 1}}{z - x}\ &=\lim_{z\rightarrow x}\frac{7z(x + 1)-7x(z + 1)}{(z + 1)(x + 1)(z - x)} \end{align*} ]

Step2: Simplify the numerator

[ \begin{align*} 7z(x + 1)-7x(z + 1)&=7zx+7z-7xz - 7x\ &=7(z - x) \end{align*} ] So, (f^{\prime}(x)=\lim_{z\rightarrow x}\frac{7(z - x)}{(z + 1)(x + 1)(z - x)})

Step3: Cancel out ((z - x))

Since (z\neq x) (as we are taking the limit (z\rightarrow x) but (z\neq 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\rightarrow 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}})