1.)y=x^5(e^{-3lnx })
The answer for this is 2x
2.) FInd an equation of the normal line to the curve $\displaystyle y = \sqrt{x^2 + 9}$ at the point $\displaystyle (4,5)$
Thanks
Note sure what this first question is asking.
The slope of the curve $\displaystyle y = \sqrt{x^2 + 9}$ at $\displaystyle x$is:2.) FInd an equation of the normal line to the curve $\displaystyle y = \sqrt{x^2 + 9}$ at the point $\displaystyle (4,5)$
$\displaystyle m(x)=\frac{d}{dx}\sqrt{x^2 + 9}=\frac{x}{\sqrt{x^2+9}}$
so when $\displaystyle x=4$ the slope is:
$\displaystyle m(4)=\frac{4}{\sqrt{16+9}}=4/5$.
The normal to the curve at $\displaystyle (4,5)$ has slope $\displaystyle -1/m(4)=-5/4$, and passess through the point, so the equation on the normal is:
$\displaystyle y=-(5/4)x+c$
where:
$\displaystyle 5=-(5/4)5+c=5+c$, so $\displaystyle c=10$, and the equation is:
$\displaystyle y=-(5/4)x+10$
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.