Question

First principle derivative of √cosecx​

Answer 1

• from first principles rule:

`{ \boxed{ \bf{ ( \delta y)/( \delta x) = {}^{lim _(x) } _(h \dashrightarrow0) \: (f(x + h) - f(x))/(h) }}}`

• f(x) → (csc x)^½

• f(x + h) → {csc (x + h)}^½

`{ \rm{ ( \delta y)/( \delta x) = {}^{lim _(x) } _(h \dashrightarrow0) \: \frac{ \{\csc(x + h) \} {}^{ (1)/(2) } - (\csc x ) {}^{ (1)/(2) } }{h} }} \\ \\ \hookrightarrow \: { \tt{rationalise : }} \\ \\ = { \rm{{}^{lim _(x) } _(h \dashrightarrow0) \: \frac{ \csc(x + h) - \csc x }{h \{ \{\csc(x + h) \} {}^{ (1)/(2) } + ( \csc x) {}^{ (1)/(2) } \} } }} \\ \\ = { \rm{{}^{lim _(x) } _(h \dashrightarrow0) \: \frac{1 - (\sin (x + h))( \csc x)}{h \{ \sin(x + h) \} \{ \csc(x + h) \} {}^{ (1)/(2) } \{ \csc x \} {}^{ (1)/(2) } }}} \\ \\ = { \rm{ = { \rm{{}^{lim _(x) } _(h \dashrightarrow0) \: \frac{1 - ( \sin(x) \cos(h) + \cos(x) \sin(h)) \csc x }{h \{ \sin(x + h) \} \{ \csc(x + h) \} {}^{ (1)/(2) } \{ \csc x \} {}^{ (1)/(2) } }}} }} \\ \\ = { \rm{ = { \rm{{}^{lim _(x) } _(h \dashrightarrow0) \: \frac{1 - \cos(h) + \cot(x) \sin(h) }{h \{ \sin(x + h) \} \{ \csc(x + h) \} {}^{ (1)/(2) } \{ \csc x \} {}^{ (1)/(2) } }}} }} \\ \\ { \tt{but : \sin(h) \approx h}} \\ { \tt{ : \cos(h) \approx 1 }} \\ \\ = { \rm{{{}^{lim _(x) } _(h \dashrightarrow0 \: ) \: \frac{h \cot(x) }{h \{ \sin(x + h) \} \{ \csc(x + h) \} {}^{ (1)/(2) } \{ \csc x \} {}^{ (1)/(2) } } }}} \\ \\ { \tt{h \: tends \: to \: 0}} \\ \\ { \boxed{ \rm{ (dy)/(dx) = - (1)/(2 √( \cot(x) \csc(x) ) ) }}}`