# Determine the equation of the line tangent to the graph of y=15e^5x/x at x=1 in the form y=mx+b has m= b=

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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

`(y-15e^5)=60e^5(x-1)`

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` .