What is the derivative of y = x^(1-e)?

Question

Also, how to find the equations of the tangent line and the normal line to the curve f(x) = x^3 - x^2 + 4 at the point (-1, 2).

I have absolutely no idea how to do this problem.

Answer 1

The trick to getting the derivative of y = x^(1-e) is to use natural logarithms before differentiating to simplify the equation.

y = x^(1-e)

ln(y) = (1-e)(ln x)

We can do this because

ln(x^y) = y (ln x)

Take the partial derivative then solve for dy/dx.

(1/y)dy = (1-e)(1/x)dx

dy/dx = (1-e)(y/x)

Substitue the original value of "y" back into the derivative and you get.

dy/dx = (1-e)( x^(1-e)/x) = (1-e)(x^(1-e-1)

dy/dx = (1-e)x^-e

The tangent line uses the slope of a curve at a specified point. All you need to do is find the equation for the derivative and plug in that value.

f(x) = x^3 - x^2 + 4
f'(x) = 3x^2 - 2x
f'(-1) = 3(-1)^2 - 2(-1) = 3 + 2 = 5 = m

So now we know the slope of the tangent line is 5. Now we use the standard form for a straight line.

y - y1 = m(x - x1)

y - 2 = 5(x - (-1)

y -2 = 5x + 1

y = 5x + 3

There is the tangent line to the point (-1,2). The normal line is the same as a perpendicular line. That means all you need to do is find the negative inverse of the slope you found.

-1/m = -1/5

Use the standard form of an equation to find the normal line.

y - 2 = (-1/5)(x - (-1))

y - 2 = (-1/5)x - 1/5

y = (-1/5)x + 9/5

edit : You're absolutely right Diane. My mind was taking a nap for a long moment. It works nonetheless and hopefully the asker will remember that exponent law for future reference. :)

This brain fart brought to you courtesy of serendipity.

Answer 2

The easier way to do it is to remember that 1-e is a constant. You know how to take the derivative of x^2 - do the same thing. The only time you need logarithmic differentiation is if you're dealing with a variable exponent.

As to the other question: to find an equation of a line, you need either two points or a point and a slope. You have one point for both the tangent and the normal, that being (-1, 2). To get the slope of the tangent, take the derivative of the curve and plug in -1 for x. Remember - the derivative is the slope of the tangent line.

The normal line is perpendicular to the tangent, so take the slope of the tangent line and take the negative reciprocal. For instance, if the slope of the tangent is 2, the slope of the normal will be -1/2.

In either case, just use the point slope form and you're done:

y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Answer 3

derivative of y=x^(1-e) is
y'=(1-e) x ^((1-e)-1)
=(1-e) x ^(-e)

f'(x)=3x^2 - 2x
f'(-1)=3(-1)^2 - 2(-1)
=3+2=5
equation: y-2=5(x+1)

Answer 4

no idea