log: Find equation of the tangent to the curve at the pt.

W

wind

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Hi, can someone check over my work? Thanks

Find the equation of the tangent to the curve at the given point

\(\displaystyle \L\ y=x^{2}lnx\) at (1,0)

\(\displaystyle \L\ y'=(2x)(lnx)+(x^{2})(\frac{1}{x})\)

\(\displaystyle \L\ y'=(2x)(lnx)+x\)

\(\displaystyle \L\ y'=(3x)(lnx)\)

\(\displaystyle \L\ y'=3ln\)

Point Slope Form

y-y1=mx(x-x1)
y-0=3 ln(x-1)
y=3 lnx-3 ln

b) y=logx, (100,2)

\(\displaystyle \L\ 10^{y}=x\)

\(\displaystyle \L\ ln10^{y}=lnx\)

\(\displaystyle \L\frac{d}{dx}ln10^{y}=\frac{d}{dx}lnx\)

\(\displaystyle \L\frac{1}{10^{y}}*{10^{y}*ln10*\frac{d}{dx}=\frac{1}{x}\)

\(\displaystyle \L\ 10^{y}*ln10*\frac{d}{dx}=\frac{1}{x}\)

\(\displaystyle \L\frac{d}{dx}=\frac{1}{(x)(10^{y})(ln10)}\)

\(\displaystyle \L\frac{d}{dx}=\frac{1}{(100)(10^{2})(ln10)}\)

\(\displaystyle \L\frac{d}{dx}=\frac{1}{(10,000)(ln10)}\)
 
y=x^2 lnx find tangent line at 1,0

slope of y is:
y'=x^2[1/x]+lnx [2x] evaluate at x=1
y'(1)=1+0

equation of tangent line is
y=mx+b but at 1,0 the slope,m,is 1
y=x+ b find b at x=1 y=0
0=1+b
b=-1

y=x-1 answer

please check for errors
 
Re: log: Find equation of the tangent to the curve at the pt

wind said:
Hi, can someone check over my work? Thanks

Find the equation of the tangent to the curve at the given point

\(\displaystyle \L\ y=x^{2}lnx\) at (1,0)

\(\displaystyle \L\ y'=(2x)(lnx)+(x^{2})(\frac{1}{x})\)

\(\displaystyle \L\ y'=(2x)(lnx)+x\)

\(\displaystyle \L\ y'=(3x)(lnx)\) what is this? 2x*lnx + x does not equal 3x*lnx ... 2x*lnx + x = x(2lnx + 1) ... y'(1) = 1(2ln1 + 1) = 1(0 + 1) = 1

\(\displaystyle \L\ y'=3ln\)

Point Slope Form

y-y1=mx(x-x1)
y-0=3 ln(x-1)
y=3 lnx-3 ln

b) y=logx, (100,2)

\(\displaystyle \L\ 10^{y}=x\)

\(\displaystyle \L\ ln10^{y}=lnx\)

y*ln(10) = ln(x)
d/dx[y*ln(10) = ln(x)]
ln(10)*(dy/dx) = 1/x
dy/dx = 1/[x*ln(10)]
at x = 100, dy/dx = 1/[100*ln(10)]
tangent line equation is ...
y - 2 = (x - 100)/[100*ln(10)]

\(\displaystyle \L\frac{d}{dx}ln10^{y}=\frac{d}{dx}lnx\)

\(\displaystyle \L\frac{1}{10^{y}}*{10^{y}*ln10*\frac{d}{dx}=\frac{1}{x}\)

\(\displaystyle \L\ 10^{y}*ln10*\frac{d}{dx}=\frac{1}{x}\)

\(\displaystyle \L\frac{d}{dx}=\frac{1}{(x)(10^{y})(ln10)}\)

\(\displaystyle \L\frac{d}{dx}=\frac{1}{(100)(10^{2})(ln10)}\)

\(\displaystyle \L\frac{d}{dx}=\frac{1}{(10,000)(ln10)}\)
Click to expand...
 
Sorry I missed this the first time

b)
find tangent line at 100,2 for function y=logx

we will find slope at point 100,2

y=logx raise both sides to power of 10
10^y=x take natural log
ln[10^y]=lnx
y ln 10 = ln x take derivative with respect to x
ln 10 dy/dx = 1/x
dy/dx = 1/[x ln10] at x=100
dy/dx=.01/ln10 this is m of tangent line

tangent line is y=mx+b at 100,2 m=.01/ln10
y=[.01/ln10] x +b at point 100,2 find b
2=[.01/ln10][100]+b
2ln10=b

y=[.01/ln10]x+2ln10 answer

please check for errors
Arthur
 
You are more than welcome. I am glad to be of help.
Please don't just copy the answers, but redo the work, for you learn from your errors.
Arthur
 
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