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A function is reprersented parametricall...

A function is reprersented parametrically by the equations `x=(1+t)/(t^3); y=3/(2t^2)+2/tdotT h e nt h ev a l u eof|(dy)/(dx)-x((dy)/(dx))^3|` is__________

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To solve the problem, we need to find the value of \(|\frac{dy}{dx} - x \left(\frac{dy}{dx}\right)^3|\) given the parametric equations: \[ x = \frac{1+t}{t^3}, \quad y = \frac{3}{2t^2} + \frac{2}{t} \] ### Step 1: Find \(\frac{dx}{dt}\) ...
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CENGAGE ENGLISH-DIFFERENTIATION-Numerical Value Type
  1. If y=f(x) is an odd differentiable function defined on (-oo,oo) such T...

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  2. If x^(3)+3x^(2)-9x+lamda is of the form (x-alpha)^(2)(x-beta) then lam...

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  3. If graph of y=f(x) is symmetrical about the point (5,0) and f^(prime)(...

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  4. Let f(x)=(x-1)(x-2)(x-3)(x-n),n in N ,a n df(n)=5040. Then the value ...

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  5. y=f(x) , where f satisfies the relation f(x+y)=2f(x)+x y(y)+ysqrt(f(x)...

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  6. If function f satisfies the relation f(x)xf^(prime)(-x)=f(-x)xf^(prime...

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  7. If y=(a+b x^(3/2))/(x^(5/4))a n dy^(prime)=0a tx=5, then the value o...

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  8. lim(hrarr0) ((e+h)^(In(e+h))-e)/(h) is-

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  9. If the function f(x)=-4e^((1-x)/2)+1+x+(x^2)/2+(x^3)/3a n dg(x)=f^(-1)...

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  10. Suppose that f(0)=0a n df^(prime)(0)=2, and let g(x)=f(-x+f(f(x)))dot ...

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  11. A nonzero polynomial with real coefficients has the property that f(x)...

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  12. A function is reprersented parametrically by the equations x=(1+t)/(t^...

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  13. Let z=(cosx)^5a n dy="sin"xdot Then the value of 2(d^2z)/(dy^2)a tx=(2...

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  14. Let g(x)={(x^2+xtanx-xtan2x)/(a x+tanx-tan3x),x!=0 0,x=0 If g^(prime)...

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  15. Let f(x)=x+1/(2x+1/(2x+1/(2x+oo))) Compute the value of f(50)dotf^(p...

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  16. Let F(x)=f(x)g(x)h(x) for all real x ,w h e r ef(x),g(x),a n dh(x) are...

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  17. "If "y=(root(3)(1+3x)root(4)(1+4x)root(5)(1+5x))/(root(7)(1+7x)root(8)...

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  18. If f(theta) = sin(tan^(-1)(sintheta/sqrt(cos2theta))), where -pi/4 lt ...

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  19. The slope of the tangent to the curve (y-x^5)^2=x(1+x^2)^2 at the poin...

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  20. Let f: R->R be a differentiable function with f(0)=1 and satisfy...

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