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
Note sure what this first question is asking.
The slope of the curve at is: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
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.
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.
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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.