# If the slope of the tangent to 4x^2+cx+2e^y=2 at x=0 is 4, then what is the value of c?

## If the slope of the tangent to $4 {x}^{2} + c x + 2 {e}^{y} = 2$ at x=0 is 4, then find c

Apr 3, 2018

The value of $c$ is $- 8$.

#### Explanation:

First we will solve for the value of $y$ at $x = 0$.

$4 {\left(0\right)}^{2} + 0 \left(x\right) + 2 {e}^{y} = 2$

$2 {e}^{y} = 2$

${e}^{y} = 1$

$y = 0$

We must find the first derivative because this gives us the slope of the tangent at $x = a$.

$8 x + c + 2 {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$2 {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - c - 8 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- c - 8 x}{2 {e}^{y}}$

We want to find at what value of $c$ that $\frac{\mathrm{dy}}{\mathrm{dx}} = 4$.

4 = (-c - 8(0))/(2(1)

$8 = - c$

$c = - 8$

Hopefully this helps!

Apr 3, 2018

$C = - 8 {e}^{y}$

#### Explanation:

$4 {x}^{2} + C x + 2 {e}^{y} = 2$, differentiating both sides , implicitly gives,

$8 \left[0\right] + C + 2 {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$, Given that the slope is $4$ at $x = 0$ we have,

$C + 8 {e}^{y}$=$0$, since $\frac{\mathrm{dy}}{\mathrm{dx}} = 4$, therefore, $C = - 8 {e}^{y}$