find an equation for a line that is normal to the graph y=e^x*x and goes through the origin.
yeah that's what i mean...but u cant plug in zero because it is not normal at the origin only goes through it..n the derivative is addition...
do we mean y = x e^x ???
at x = 0, y = 0 so it goes through the origin
slope = dy/dx = x e^x - e^x
at x = 0 that is e^0 = 1
so our slope = -1/1 = -1
so our line is
y = -1 x + 0
or
y = -x
sure it is normal. slope = -1/slope of function at origin
Maybe you did not follow the derivative? I used the product rule
d/dx(uv) = u dv/dx + v du/dx
here u = x
and v = e^x
so
du/dx = 1
dv/dx = e^x
so
df/dx = x e^x + 1 e^x
at zero that is one because e^0 is one
so our function slope at the origin is m = 1
normal slope = -1/m = -1/1 = -1
To find the equation for a line that is normal to the graph of y = e^x*x and goes through the origin, we need to determine the slope of the line.
First, let's find the derivative of the given function y = e^x*x. Taking the derivative with respect to x, we get:
dy/dx = (d/dx)(e^x*x)
= e^x * x + e^x * 1
= e^x * (x + 1)
The slope of the normal line will be the negative reciprocal of the derivative at the point of tangency. Since the line goes through the origin (0, 0), the slope of the normal line is simply the negative reciprocal of dy/dx at x = 0.
Substituting x = 0 into the derivative expression, we get:
dy/dx = e^0 * (0 + 1)
= 1
Therefore, the slope of the normal line is -1 (negative reciprocal of 1). So, the normal line has a slope of -1.
Now, we can use the point-slope form of the equation of a line to find the equation. Since the line goes through the origin (0, 0) and has a slope of -1, we can use the equation:
y - y1 = m(x - x1)
where (x1, y1) is the point on the line (in this case, the origin), and m is the slope. Substituting the values, we get:
y - 0 = -1(x - 0)
y = -x
Therefore, the equation for the line that is normal to the graph y = e^x*x and goes through the origin is y = -x.