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 4x^2+cx+2e^y=2 at x=0 is 4, then find c

2 Answers
Apr 3, 2018

The value of c is -8.

Explanation:

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

4(0)^2 +0(x) + 2e^y =2

2e^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.

8x + c + 2e^y(dy/dx) = 0

2e^y(dy/dx) = -c - 8x

dy/dx= (-c - 8x)/(2e^y)

We want to find at what value of c that dy/dx= 4.

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

8 = -c

c = -8

Hopefully this helps!

Apr 3, 2018

C=-8e^y

Explanation:

4x^2+Cx+2e^y=2, differentiating both sides , implicitly gives,

8[0]+C+2e^ydy/dx=0, Given that the slope is 4 at x=0 we have,

C+8e^y=0, since dy/dx=4, therefore, C=-8e^y