not quite
if f(x) = x
f'(x) = 1
so its
y' = 4e^x
so you you function the derivative is
\[y' = 4x^3 + 4e^x\]
ok... you now have the equation for the slop of the tangent...(1st derivative)
subsitiute x = 0 to find the slope of the tangent at the point (0, 4)
thats correct so you have a tangent or line with a slope of 4 and point (0, 4)
can you find the equation of the line...?
for the normal... the slope will be the negative reciprocal of the slope of the tangent... what would the slope of the normal be..?
no... ok... lets step back...
you have a slope of 4 and point (0, 4)
the equation of the tangent will be
y = 4x + b
or using the formula
\[y - 4 = 4(x - 0)\]
whats the equation of the tangent...?
ok... thats the equation of the tangent...
now the normal... which is perpendicular to the tangent...
the slope is the negative reciprocal of the tangent...
so if the tangent has a slope of 4 ... what is the slope of the normal..?