1.)y=x^5(e^{-3lnx })

The answer for this is 2x

2.) FInd an equation of the normal line to the curve at the point

Thanks :) :)

Printable View

- January 21st 2007, 08:44 PM^_^Engineer_Adam^_^PLease Differentiate...
1.)y=x^5(e^{-3lnx })

The answer for this is 2x

2.) FInd an equation of the normal line to the curve at the point

Thanks :) :) - January 21st 2007, 09:36 PMCaptainBlack
Note sure what this first question is asking.

Quote:

2.) FInd an equation of the normal line to the curve at the point

so when the slope is:

.

The normal to the curve at has slope , and passess through the point, so the equation on the normal is:

where:

, so , and the equation is:

RonL - January 21st 2007, 10:34 PMAfterShock
I think he wants help differentiating it.

y = x^5*e^(-3*ln(x))

dy/dx = 5x^4*e^(-3*ln(x)) + x^5*(deriv. of e^(-3*ln(x)) here)

-3*1/x*e^(-3*ln(x)) = -3/x*e^(-3*ln(x)) = (-3/x)*(1/x^3) = -3/x^2

Thus, we have:

5x^4*e^(-3*ln(x)) + x^5*(-3/x^2)

5x^4*e^(-3*ln(x)) - 3x

5x^4*1/(x^3) - 3x

(5x^4)/(x^3) - 3x

5x - 3x =**2x**, which agrees with the answer you thought it might be. - January 21st 2007, 11:12 PMearboth
- January 21st 2007, 11:22 PMticbol
Here is one way.

1) y = (x^5)[e^(-3ln(x))]

Differentiate both sides with respect to x.

Let us evaluate first the e^(-3ln(x)).

Let e^(-3ln(x)) = N

Take the ln of both sides,

(-3ln(x))ln(e) = ln(N)

(-3ln(x))(1) = ln(N)

-3ln(x) = ln(N)

ln(x^(-3)) = ln(N)

Hence,

N = x^(-3)

So, e^(-3ln(x)) = x^(-3) --------***

That's very long, I know, but I don't know if you know that

e^(ln x^(-3)) = x^(-3).

So,

y = (x^5)[e^(-3ln(x))]

y = (x^5)(x^(-3))

y = x^2

dy/dx = 2x --------------answer.

----------------------------------------------------

2) 2.) FInd an equation of the normal line to the curve y = sqrt(x^2 +9) at the point (4,5)

dy/dx is the slope of the tangent line.

The slope of the normal line is the negative reciprocal dy/dx, which is -dx/dy.

y = sqrt(x^2 +9)

Square both sides,

y^2 = x^2 +9

x = sqrt(y^2 -9)

dx/dy = (1/2)[(y^2 -9)^(-1/2)](2y)

dx/dy = y /sqrt(y^2 -9)

At point (4,5),

dx/dy = 5 /sqrt(5^2 -9) = 5 /sqrt(16) = 5/4

Hence, slope of normal line = -5/4.

Slope = -5/4

Point(4,5)

Point-slope form of the equation of the line,

(y -y1) = m(x -x1)

(y -5) = (-5/4)(x -4)

Clear the fraction, multiply both sides by 4,

4y -20 = -5x +20

5x +4y -40 = 0 -------------------the normal line, in standard form

In y = mx +b form,

5x +4y -40 = 0

4y = -5x +40

y = -(5/4)x +10 -------------the normal line.