how to solve this calculus problem?
the slope of the line normal to the curve y=xe^sinx at x=2.438
- aapibushraLv 58 years ago
dy/dy = x.cosx.e^sinx + e^sinx (using product rule here)
dy/dx = (2.438).cos(2.438).e^sin(2.438) + e^sin(2.438)
dy/dx = 3.585
But this is the gradient of the curve at x=2.438
The gradient of the normal to the curve = -1/3.585
- GuillermoLv 78 years ago
Download Graph 4.4 from www.padowan.dk for free.
On "Function I Insert function", type "x*exp(sin(x))", then "OK".
Click on the 7th icon, type "2.438", then "OK".
The normal line is 1.6405*x + 8.6555; the slope is 1.6405, answer!Source(s): Graph 4.4