Use the quotient rule to differentiate both functions and then use the difference rule.
Let #g(x) = e^x/x#
#g'(x) = (e^x xx x - e^x xx 1)/x^2#
#g'(x) = (xe^x - e^x)/x^2#
#g'(x) = (e^x(x - 1))/x^2#
#:.# Let #h(x) = e^(2x - 3)/x^3#
We must differentiate the numerator using the chain rule--
Let #y = e^u#
#u = 2x - 3#
#y' = e^u#
#u' = 2#
#:.y' = e^(2x - 3) xx 2#
As for the denominator:
#y = x^3#
#y' = 3x^2#
Hence, #h'(x) = (2e^(2x - 3) xx x^3 -e^(2x - 3) xx 3x^2)/(x^3)^2#
#h'(x) = (2x^3e^(2x - 3) - 3x^2e^(2x -3))/x^6#
#h'(x) = (e^(2x - 3)(2x^3 - 3x^2))/x^6#
#:. f'(x) = (e^x(x - 1))/x^2 - (e^(2x - 3)(2x^3 - 3x^2))/x^6#
The slope of the tangent is given by evaluating #f'(a)#, a being the given point, #x = a#.
So, we have
#f'(2) = (e^2(2 - 1))/2^2 - (e^(2(2) - 3)(2(2)^3 - 3(2)^2))/2^6#
#f'(2) = e^2/4 - (4e)/64#
#f'(2) = e^2/4 - e/16#
#f'(2) = (4e^2 - e)/16#
The slope of the tangent is #(4e^2 - e)/16#.
We need to find the y value that the function and the tangent passes through.
#f(2) = e^2/2 - e^1/8#
#f(2) = (4e^2 - e)/8#
Now that we know a point on the function and on the tangent and the slope of the tangent, we can find the equation of the tangent.
#y - y_1 = m(x - x_1)#
#y - ((4e^2 - e)/8) = (4e^2 - e)/16(x - 2)#
#y - (4e^2 + e)/8 = (4e^2x - 8e^2 - ex + 2e)/16#
#y = (4e^2x - 8e^2 - ex + 2e)/16 + (4e^2 + e)/8#
#y = (4e^2x - 8e^2 - ex + 2e + 8e^2 + 2e)/16#
#y = 1/16e(4ex - x + 4)#
Thus, the equation of the tangent is #y = 1/16e(4ex - x + 4)#.
Practice exercises:
- Determine the equation of the tangent.
a) #y = sqrt(3x + 1)#, at #x = 5#
b) #y = e^(2x)/x^3 + e^(x^2 - 1)/x#, at #x = 1#
Hopefully this helps, and good luck!
