Equation of the line tangent.. please help work shown!

[johnq2k7]

Give the equation of the line tangent to the curve at the given point

a.) y*tan^-1 (x)= x*y at (sqrt(3),0)

b.) ln y= x^2 +2*e^x at (0,e^2)

work process:

a.) y*tan^-1 (x) - x*y=0
y( tan^-1(x) - x)=0
y= 0/ (tan^-1(x) -x)
y=0

also, y*tan^-1(x)= x*y becomes tan^-1(x)=x

since this is only possible if x= 0, since tan^-1(x) will not equal x at any other value

at point (sqrt(3),0) tan^-1(sqrt(3) is not equal to sqrt(3)

so the curve at the the point (sqrt(3),0) is the line y=0 so it's tangent is line y=0 as well

i'm totally lost for that question.... please help


b.) ln y = x^2 +2*e*^x at (0,e^2)

since e^(ln x) = x

therefore y= e^(x^2) + e^(2*e^x)

 

[johnq2k7]

part b continued:

dy/dx= 2*x*e^(x^2) + e^(2*e^x)*(2*e^x) = 2*x*e^x +2*e^(2*e^x+x)= 2(x*e^x+ e^(2*e^x +x))

since point line equation is y-y1= m(x-x1)

therefore, y- e^2= (2x*e^x +e^(2e^x +x)) (x-0)

therefore equation of tangent to the curve is y= 2x^2*e^x + xe^(2e^x+ x) + e^2

is this correction?

 

[Deleted member 4993]

Give the equation of the line tangent to the curve at the given point

a.) y*tan^-1 (x)= x*y at (sqrt(3),0)

b.) ln y= x^2 +2*e^x at (0,e^2)

work process:

a.) y*tan^-1 (x) - x*y=0
y( tan^-1(x) - x)=0
y= 0/ (tan^-1(x) -x)
y=0

also, y*tan^-1(x)= x*y becomes tan^-1(x)=x

since this is only possible if x= 0, since tan^-1(x) will not equal x at any other value

at point (sqrt(3),0) tan^-1(sqrt(3) is not equal to sqrt(3)

so the curve at the the point (sqrt(3),0) is the line y=0 so it's tangent is line y=0 as well

i'm totally lost for that question.... please help

I think there is something seriously wrong with this question . The curve basically is y-axis (line x=0) and x-axis (line y=0). The given point is not on the sets of solutions.

b.) ln y = x^2 +2*e*^x at (0,e^2)

since e^(ln x) = x

therefore y= e^(x^2) + e^(2*e^x) <<< no need to do that

use implicit differentiation

y'/y = 2x + 2 * e^x

So now you know the slope of the tangent line and a point on the line.

Find the equation....