#Differentiation
11 messages · Page 1 of 1 (latest)
a) Chain rule + product rule
b) OA×OB×½=Area
a)
||Let f(x)=xe^(x²)
Equation of tangent to f(x) at x=1:
y=f'(1)(x-1)+f(1)
f(1)=e
f'(x) = 2x²e^(x²) + e^(x²)
f'(1)=2e+e=3e
→ tangent equation:
y=3e(x-1)+e
y=3ex-2e||
b)
||y=3ex-2e → B=-2e, A=⅔
½×⅔×2e=2e/3||
for part A: use product rule v du/ dx + u dv/dx
where u could be x and v could be e^x
get an expression for dy/dx
and then sub in x = 1 to find the gradient at that point
to find y coordinates sub x = 1 into y = xe^x
use formula: y-y1 = m(x-x1) to find equation of line where m is the gradient at x = 1, x = 1 and y = e