if-x-2-3xy-y-2-3-find-dy-dx-at-point-1-1-hence-differentiate-sin-x-1-x-with-respect-to-x-

Question Number 38252 by Rio Mike last updated on 23/Jun/18

i f x^{2} + 3 x y - y^{2} = 3 f i n d

\frac{d y}{d x} a t p o i n t (1, 1) h e n c e

d i f f e r e n t i a t e \frac{s i n x}{1 + x} w i t h r e s p e c t

t o x .

Commented by math khazana by abdo last updated on 23/Jun/18

x^{2} + 3 x y - y^{2} = 3 \Rightarrow y^{2} - 3 x y - x^{2} + 3 = 0

Δ = 9 x^{2} - 4 (3 - x^{2}) = 13 x^{2} - 12 s o i f x^{2} ⩾ \frac{12}{13}

y (x) = \frac{3 x \bar{+} \sqrt{13 x^{2} - 12}}{2} \Rightarrow

y^{^{'}} (x) = \frac{3}{2} \bar{+} \frac{1}{2} \frac{26 x}{2 \sqrt{13 x^{2} - 12}}

= \frac{3}{2} \bar{+} \frac{13}{2 \sqrt{13^{} x^{2} - 12}} \Rightarrow

y^{^{'}} (1) = \frac{3}{2} \bar{+} \frac{13}{2} \Rightarrow y^{^{'}} (1) = 8 o r y^{^{'}} (1) = - 5

Commented by math khazana by abdo last updated on 23/Jun/18

l e t f (x) = \frac{s i n x}{1 + x} s o i f x \neq - 1

f^{^{'}} (x) = \frac{c o s x (1 + x) - s i n x}{{(1 + x)}^{2}} = \frac{c o s x + x c o s x - s i n x}{{(1 + x)}^{2}}