# 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=

rakesh05 | High School Teacher | (Level 1) Assistant Educator

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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` .