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TARGET PUBLICATION-DIFFERENTIATION -DERIVATIVE OF INVERSE FUNCTIONS
- If y = sin^(-1) (19/20x) + cos^(-1)(19/20x), then (dy)/(dx)=
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- I fy=sec^(- 1)((x+1)/(x-1))+sin^(- 1)((x-1)/(x+1)),t h e n(dy)/(dx)
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- If y=sec^-1([sqrtx+1]/[sqrtx-1])+sin^-1([sqrtx-1]/[sqrtx+1]), then dy/...
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- (d)/(dx)[sin^(-1)x+sin^(-1)sqrt(1-x^(2))]=
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- d/(dx)[sin {2 cos^(-1) (sin x)}]=
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- Find the differentiation of y=tan^(-1)((x^(1//3)+a^(1//3))/(1-x^(1//3)...
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- If y=tan^(-1)((6+5 tanx)/(5- 6 tanx)), then (dy)/(dx)=
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- Find (dy)/(dx) if y=tan^(-1)(4x)/(1+5x^2)+tan^(-1)(2+3x)/(3-2x)
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- Find (dy)/(dx) if y=tan^(-1)(4x)/(1+5x^2)+tan^(-1)(2+3x)/(3-2x)
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- d/(dx)(tan^(-1) ((cos x)/(1+sinx))=
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- If y = tan^(-1) (sec x - tan x ) , "then" (dy)/(dx) is equal to
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- Differentiate w.r.t. x: (i)tan^(-1){sqrt((1+cosx)/(1-cosx))}" "(...
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- If coty=(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt1-sinx)," then "...
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- If y= sin ^(-1) ((4cos x +5sin x )/( sqrt(41))) ,then (dy)/(dx)=
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- If y=tan^(-1) ((1+x)/(1-x)) then (dy)/(dx)=
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- d/(dx)(tan^(- 1)(2/(x^(- 1)-x))) is equal to
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- d/dx(tan^-1(x/sqrt(a^2-x^2))
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- d/dx(tan^-1(x/sqrt(a^2-x^2))
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- If y=sin^(-1)((2x)/(1+x^2))+sec^(-1)((1+x^2)/(1-x^2)),-<x<1, Prove tha...
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- (d)/(dx)[sin^(-1)sqrt(((1-x))/(2)]=
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