Question

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

Answer 1

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!

Answer 2

`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`