# if y= 1+(a/x-a)+(bx/(x-a)(x-b))+(cx^2/(x-a)(x-b)(x-c)) than show that dy/dx=y/x[(a/a-x)+(b/b-x)+(c/c-x)].............

SAGAR SINGH - IIT DELHI
879 Points
12 years ago

[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)

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Sagar Singh

B.Tech, IIT Delhi

509 Points
12 years ago

y = 1 + a/(x-a) +bx/(x-a)(x-b) + cx2/(x-a)(x-b)(x-c)

breaking into different  functions

y = 1 +    f(x)1  +       f(x)2       +        f(x)3

f(x)1=a/(x-a)    ,        f(x)2 =bx/(x-a)(x-b)   &   f(x)3 =cx2 /(x-a)(x-b)(x-c)

now

dy/dx= d/dx(1) + d/dx(f(x)1) +d/dx(f(x)2) +d/dx(f(x)3) ....................................1

now separatly differentiatiating f(x)1 , f(x)2  , f(x)3  by taking log

d/dx[log(f(x)1)]  =  d/dx(loga/x-a)

d/dx(f(x)1/f(x)1 = d/dx(loga - log(x-a) )

d/dx(f(x)1)=(-1/x-a)(f(x)1

d/dx(f(x)1) =    -a/(x-a)2

now separatly differentiate each of these and put in eq 1 ,u will get the required result