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

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

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

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Answer 1 · 2018-04-03T18:54:50

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

Explanation:

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

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

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

$e^{y} = 1$ $e^y = 1$

$y = 0$ $y = 0$

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

$8 x + c + 2 e^{y} (\frac{d y}{d x}) = 0$ $8x + c + 2e^y(dy/dx) = 0$

$2 e^{y} (\frac{d y}{d x}) = - c - 8 x$ $2e^y(dy/dx) = -c - 8x$

$\frac{d y}{d x} = \frac{- c - 8 x}{2 e^{y}}$ $dy/dx= (-c - 8x)/(2e^y)$

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

$4 = \frac{- c - 8 (0)}{2 (1)}$ $4 = (-c - 8(0))/(2(1)$

$8 = - c$ $8 = -c$

$c = - 8$ $c = -8$

Hopefully this helps!

Answer 2 · 2018-04-03T21:56:05

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

Explanation:

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

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

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

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

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

2 Answers

Explanation:

Explanation:

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