Question

please answer this question​

Answer 1

Answer:

The derivative of $\tt{ \dfrac{1}{1-x} }$ is $\tt{ \dfrac{1}{(1-x)^{2}} }$.

Step-by-step explanation:

w.k.t., derivative of a function f(x) using first principle is $\tt{ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} }$

Let f(x) = $\tt{ \dfrac{1}{1-x} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{\dfrac{1}{1-(x+h)}-\dfrac{1}{1-x}}{h} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{\dfrac{1}{1-x-h}-\dfrac{1}{1-x}}{h} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{1-x-(1-x-h)}{h(1-x-h)(1-x)} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{1-x-1+x+h}{h(1-x-h)(1-x)} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{h}{h(1-x-h)(1-x)} }$

$\tt{ \implies f'(x) = \lim_{h \to 0} \dfrac{1}{(1-x-h)(1-x)} }$

$\tt{ \implies f'(x) = \dfrac{1}{(1-x-0)(1-x)} }$

$\tt{ \implies f'(x) = \dfrac{1}{(1-x)(1-x)} }$

$\tt{ \implies f'(x) = \dfrac{1}{(1-x)^{2}} }$