Question

If given that :-

${ \quad \leadsto \quad { \sf y = 1 + \dfrac{\alpha}{\bigg( \dfrac{1}{x} - \alpha \bigg)} + \dfrac{\bigg(\dfrac{\beta}{x} \bigg)}{\bigg( \dfrac{1}{x} - \alpha \bigg) . \bigg( \dfrac{1}{x} - \beta \bigg)} + \dfrac{\bigg( \dfrac{\gamma}{x²} \bigg)}{\bigg( \dfrac{1}{x} - \alpha \bigg) . \bigg( \dfrac{1}{x} - \beta \bigg) . \bigg ( \dfrac{1}{x} - \gamma \bigg)} }}$

Find the derivative of ' y ' w.r.t.x from both ab - initio method & simple differentiation .​

Answer 1

Answer:

[a/(x-a)]+1=x/(x-a)

[bx/(x-b)(x-c)]+x/(x-a)=x/(x-a)[b/(x-b)+1]=x2/(x-a)(x-b)

So y={cx^2/ (x-a) (x-b) (x-c)} +x2/(x-a)(x-b)=x2/((x-a)(x-c))[c/(x-c)+1]

y=x3/(x-a)(x-b)(x-c)

logy=3logx-log(x-a)-log(x-b)-log(x-c)

y1/y=3/x+1/(a-x)+1/(b-x)+1/(c-x)

y1/y=[1/x-1/(x-a)]+[1/x-1/x-b]+[1/x-1/x-c]=1/x ( a/a-x + b/ b-x+ c/c-x)

Answer 2

Answer:

${ \quad \leadsto \quad { \sf y = 1 + \dfrac{\alpha}{\bigg( \dfrac{1}{x} - \alpha \bigg)} + \dfrac{\bigg(\dfrac{\beta}{x} \bigg)}{\bigg( \dfrac{1}{x} - \alpha \bigg) . \bigg( \dfrac{1}{x} - \beta \bigg)} + \dfrac{\bigg( \dfrac{\gamma}{x²} \bigg)}{\bigg( \dfrac{1}{x} - \alpha \bigg) . \bigg( \dfrac{1}{x} - \beta \bigg) . \bigg ( \dfrac{1}{x} - \gamma \bigg)} }}$

Step-by-step explanation:

${ \quad \leadsto \quad { \sf y = 1 + \dfrac{\alpha}{\bigg( \dfrac{1}{x} - \alpha \bigg)} + \dfrac{\bigg(\dfrac{\beta}{x} \bigg)}{\bigg( \dfrac{1}{x} - \alpha \bigg) . \bigg( \dfrac{1}{x} - \beta \bigg)} + \dfrac{\bigg( \dfrac{\gamma}{x²} \bigg)}{\bigg( \dfrac{1}{x} - \alpha \bigg) . \bigg( \dfrac{1}{x} - \beta \bigg) . \bigg ( \dfrac{1}{x} - \gamma \bigg)} }}$

◆Solution:

~~Finding new values.