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Determine the equation of the line tangent to the graph of y=15e^5x/x at x=1 in the...
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High School Teacher
Given the function `y=(15e^(5x))/x` .
We know that equation of tangent to the curve y=f(x) at `(x_1,y_1)` is given by
`y-y_1=m(x-x_1)` where `m=dy/dx` at `(x_1,y_1)` .
Here `x_1=1` . So for `x_1=1, y_1=(15e^(5.1))/1=15e^5` .
Now `dy/dx=(15e^(5x)(5x-1))/x^2` .
So, `dy/dx` at `(1,15e^5)` =60`e^5` . So, `m=60e^5` .
Now equation of the tangent is
or, `y=60e^5x-45e^5` .
This is the reqired equation of tangent to the graph of `y=(15e^(5x))/x` at x=1 in the desired form.
Here if we compare the above equation of tangent with `y=mx+b` we get
`m=60e^5` and `b=-45e^5` .
Posted by rakesh05 on May 14, 2013 at 6:17 AM (Answer #1)
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