determine the equation of the line tangent to the graph of y=15e^(4x)/ x at x=1 in the form y=mx +b has m= b=
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
You should remember what is the equation of the tangent line to the graph of the function, at a point `x =x_0` such that:
`y - f(x_0) = f'(x_0)(x - x_0)`
You should find the derivative of the function using the quotient rule and the chain rule, such that:
`(dy)/(dx) = f'(x) = ((15e^(4x))'*x - 15e^(4x)*(x)')/x^2`
`f'(x) =(15*4x*e^(4x) - 15e^(4x))/x^2`
Factoring out `15e^(4x)` such that:
`f'(x) = (15e^(4x)(4x - 1))/x^2`
You need to find f'(1), hence, you should substitute 1 for x in equation `f'(x) = (15e^(4x)(4x - 1))/x^2` such that:
`f'(1) = (15e^(4)(4 - 1))/1^2 => f'(1) = 45e^4`
`f(1) = 15e^4`
Hence, you may write the equation of the tangent line such that:
`y - 15e^4 = 45e^4(x - 1)`
You need to put the equation `y - 15e^4 = 45e^4(x - 1)` in the slope intercept form such that:
`y = 15e^4 + 45e^4*x - 45e^4`
`y = 45e^4*x - 30e^4`
Hence, evaluating the equation of the tangent line yields `y = 45e^4*x - 30e^4` , where `m = 45e^4` and `b = - 30e^4` .
Related Questions
- Find the equation of the tangent line to the curve y=5sec(x)-10cos(x) at the point...
- 1 Educator Answer
- Use linear approximation, i.e. the tangent line, to approximate 125.1^(1/3) as follows: Let...
- 1 Educator Answer
- How do I set up this equation in y=mx+b form?Please explain how to set up this equation in y=mx+b...
- 1 Educator Answer
- Find the equation of the tangent line to the curve y=5xcosx at the point (pi,-5pi).The equation...
- 1 Educator Answer
- determine the equation of the line tangent to the graph of y=9e^(3x^2)-4x at x=3 in the form...
- 1 Educator Answer
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.