Question

Find dy/dx and d2y/dx2. x = t2 + 5, y = t2 + 5t dy dx = Correct: Your answer is correct. d2y dx2 = Correct: Your answer is correct. For which values of t is the curve concave upward? (Enter your answer using interval notation.) Changed: Your submitted answer was incorrect. Your current answer has not been submitted.

Answer 1

Answer:

The answer to this question can be defined as follows:

Explanation:

Given value:

$x = t^2 + 5......(1)\\\\ y = t^2 + 5t........(2)$

To find:

$\bold {\frac{dy}{dx} \ \ \ and\ \ \ \frac{d^2y}{dx^2} = ?}$

Differentiate the above equation:

equation 1:

$\frac{dx}{dt}= 2t.......(1)$

equation 2:

$\frac{dy}{dt}= 2t+5$

Formula:

$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\boxed{\bold{\frac{dy}{dx}=\frac{2t+5}{2t}}}$

To calculate the $\bold{\frac{d^2y}{dx^2}}$ we Differentiate the above equation but before that first solve the equation:

Equation:

$\frac{dy}{dx}=\frac{2t+5}{2t}$

    $=\frac{2t}{2t}+\frac{5}{2t}\\\\= 1+\frac{5}{2t}\\\\=1+\frac{5}{2} t^{-1}$

Formula:

$\bold{\frac{d}{dx} \ x^n = nx^{n-1}}$

$\frac{dy^2}{dx^2}= 0+\frac{5}{2} (-1 t^{-2})$

      $= -\frac{5}{2} t^{-2}\\\\= -\frac{5}{2 t^2}$

$\boxed{\bold{\frac{d^2y}{dx^2}=-\frac{5}{2t^2}}}$

Answer 2

Answer:

d2y dx2

Explanation: