am i correct?? since im really bad with trigo
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Answer:
Calculate the equation(s):
y = e^-x, A(-1, e)
Use the definition y = f(x) to replace y.
f(x) = e^-x
Take the derivative of both sides.
d/dx(f(x)) = d/dx(e^-x)
Use the substitution u = e^-x and the Chain Rule to evaluate the derivative.
f'(x) = u × d/dx(-x)
The derivative of a variable to the first power is always 1.
f'(x) = e^-x × -1
Multiplying by -1 changes the sign of the expression.
f'(x) = -e^-x
Substitute x as -1 using the given point A to find the slope of the tangent.
f'(x) = -e^-(-1)
Calculate the product.
f'(x) = -e^1
Any number raised to the first power will be equal to that number.
f'(x) = -e
Using the equation of the tangent y - y1 = f'(x)(x - x1), substitute all given terms:
y - e = -e(x - (-1))
Calculate the product.
y - e = -e(x + 1)
Move -e to the right-hand side and change its sign.
y = -e(x + 1) + e
Distribute -e through the parentheses.
y = -ex - e + e
Eliminate the opposites.
y = -ex
To find the equation of the normal, use the equation y - y1 = -1/f'(x)(x - x1).
y - e = -1/e(x - (-1))
Simplify.
y = -1/e(x + 1) + e
Distribute -1/e through the parentheses.
y = -x/e - 1/e + e
Write all denominators above the LCD e.
y = (-x - 1 + e²) / e
Use the commutative property to rearrange the terms.
y = (-x + e² - 1) / e