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TARGET PUBLICATION-DIFFERENTIATION -COMPETITIVE THINKING
- If y = f(x^(2) + 2) " and " f'(3) = 5 , " then" dy/dx " at x " = 1 is
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- If f (x) = log(x) (log x), then f'(x) at x = e is …….. .
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- If f(x)=sqrt(1+cos^2(x^2)),thenf'(sqrtpi/2) is
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- If : f(x)=(sin^(2)x)/(1+cotx)+(cos^(2)x)/(1+tanx)," then: "f'((pi)/(4)...
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- d/(dx)tan^-1((1-x)/(1+x))=
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- If y =tan ^(-1) ((sqrt( a) -sqrt(x)) /( 1+sqrt( ax)) ) ,then (dy)/(dx)...
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- If y=tan^(-1)[(sinx-cosx)/(cosx-sinx)] then (dy)/(dx) is
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- If y= tan ^(-1) ((acos x -bsin x )/( bcos x+asin x ) ) ,then (dy)/(dx...
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- यदि y= sec tan ^(-1) x, तब (dy)/(dx) =
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- the derivative of tan^-1 (6xsqrtx)/(1-9x^3) is sqrtx g(x) then g(x) is...
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- If y= e^(m sin ^(-1)) x and (1-x^(2))((dy)/(dx))^(2)= Ay^(2). then ...
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- If y=sin^(-1)[x(1-x)-sqrtxsqrt(1-x^(2))], find (dy)/(dx),
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- Find (dy)/(dx) when(y-tan^(- 1))x/(1+sqrt(1-x^2))+sin[2tan^(- 1)sqrt((...
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- If f(x)=cot^(-1)((x^(x)-x^(-x))/(2)) then f'(1) equals
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- What is the derivative of tan^(-1)((sqrtx-x)/(1+x^(3//2))) at x = 1?
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- If y=(tan^- 1)(sqrt(1+x^2)-1)/x, then y'(1) is equal to
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- If y=(1+1/x)^x,then(dy)/(dx)=
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- If y=(sinx)^(tanx),then(dy)/(dx) is equal to
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- If y=(e^(2x)cosx)/(x sinx),then(dy)/(dx)=
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- If y={f(x)}^(phi(x)),"then"(dy)/(dx) is
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