Application of ...

The slope of the tangent to the curve $$y=\displaystyle\int_{0}^{x}\dfrac{dt}{1+t^3}$$ at the point where x=1 is 

Consider the curve $$y = e^{2x}$$.What is the slope of the tangent to the curve at (0, 1) ?

If a tangent to the curve $$\displaystyle y=6x-{ x }^{ 2 }$$ is parallel to the line $$\displaystyle 4x-2y-1=0$$, then the point of tangency on the curve is:

The function $$x^{x}$$ is increasing, when

The slope of the tangent to the curve $$y = \int_{0}^{x} \dfrac {dt}{1 + t^{3}}$$ at the point where $$x = 1$$ is

If tangent to the curve $$\displaystyle x={ at }^{ 2 },y=2at$$ is perpendicular to $$x$$-axis, then its point of contact is:

Equation of normal drawn to the graph of the function defined as $$f\left( x \right) =\dfrac { \sin { { x }^{ 2 } }  }{ x } ,x\neq 0$$ and $$f\left( 0 \right) =0$$ at the origin is

The gradient of the tangent line at the point $$(a cos \alpha, a sin \alpha)$$ to the circle $$x^2 + y^2 = a^2$$, is

Consider the following statements:
1. $$\dfrac {dy}{dx}$$ at a point on the curve gives slope of the tangent at that point.
2. If $$a(t)$$ denotes acceleration of a particle, then $$\displaystyle \int a(t) dt + c$$ give velocity of the particle.
3. If $$s(t)$$ gives displacement of a particle at time $$t$$, then $$\dfrac {ds}{dt}$$ gives its acceleration at that instant.
Which of the above statements is/ are correct?

If the product of the slope of tangent to curve at $$(x,y)$$ and its y-co-ordinate is equal to the x-co-ordinate of the point, then it represent.