Stationary Points of a Function.

1.	Stationary Points of a Function.
Answer» y = f(x). Points at which the maximum and minimum values of f(x) are obtained are called the stationary points of function. Necessary condition for stationary points: f'(x) = 0

Discussion

`"If "y^(1//m)=(x+sqrt(1+x^(2)))," then "(1+x^(2))y_(2)+xy_(1)` is (where `y_(r)` represents the rth derivative of y w.r.t. x)A. `m^(2)y`B. `my^(2)`C. `m^(2)y^(2)`D. none of these
Find `(dy)/(dx)`, when: `y=e^(x)sin^(3)xcos^(4)x`
If y= sin px and `y_(n)` is the nth derivative of y, then `|{:(y,y_(1),y_(2)),(y_(3),y_(4),y_(5)),(y_(6),y_(7),y_(8)):}|`isA. 1B. 0C. -1D. none of these
Find `(dy)/(dx)`, when: `y=2^(x).e^(3x)sin4x`
Find the second derivative of `e^(3x)sin4x.`
If for `x (0,1/4),`the derivative of `tan^(-1)((6xsqrt(x))/(1-9x^3))`is `sqrt(x)dotg(x),`then `g(x)`equals:`(3x)/(1-9x^3)`(2) `3/(1+9x^3)`(3) `9/(1+9x^3)`(4) `(3xsqrt(x))/(1-9x^3)`A. `(9)/(1+9x^(3))`B. `(3xsqrt(x))/(1-9x^(3))`C. `(3x)/(1-9x^(3))`D. `(3)/(1+9x^(3))`
The slope of the tangent to the curve `(y-x^5)^2=x(1+x^2)^2`at the point `(1,3)`is.
If for `x (0,1/4),`the derivative of `tan^(-1)((6xsqrt(x))/(1-9x^3))`is `sqrt(x)dotg(x),`then `g(x)`equals:`(3x)/(1-9x^3)`(2) `3/(1+9x^3)`(3) `9/(1+9x^3)`(4) `(3xsqrt(x))/(1-9x^3)`A. `(3)/(1+9x^(3))`B. `(9)/(1+9x^(3))`C. `(3xsqrt(x))/(1-9x^(3))`D. `(3x)/(1-9x^(3))`
If y=c`e^(x//(x-a))`, then `(dy)/(dx)` equalsA. `a(x-a)^(2)`B. `-(ay)/((x-a)^(2))`C. `a^(2)(x-a)^(2)`D. `a(x-a)`
If `f(theta) = sin(tan^(-1)(sintheta/sqrt(cos2theta)))`, where `-pi/4 lt theta lt pi/4,` then the value of `d/(d(tantheta)) f(theta)` is