Asked by calc
Find an equation of the tangent line to the curve at the given point.
y = 6 x sin x
P= (pi/2 , 3pi)
i know the slope of a tangent line is equal to the first derivative. For that I got 6xcosx + 6sinx but idk how to put that into the y-y1=m(x-x1) formula to make the equation
y = 6 x sin x
P= (pi/2 , 3pi)
i know the slope of a tangent line is equal to the first derivative. For that I got 6xcosx + 6sinx but idk how to put that into the y-y1=m(x-x1) formula to make the equation
Answers
Answered by
TutorCat
great job on the first derivative!
m=slope
m= 6(pi/2)cos(pi/2) + 6sin(pi/2)
y1=3pi
x1=pi/2
m=slope
m= 6(pi/2)cos(pi/2) + 6sin(pi/2)
y1=3pi
x1=pi/2
Answered by
calc
oh! ok! thank you!
so then m=6 and the equation of the tangent line would be y=6x
Thank you!
so then m=6 and the equation of the tangent line would be y=6x
Thank you!
Answered by
TutorCat
remember to plug in your y1 and x1 & solve for y!
y-y1=m(x-x1)
y-3pi=6(x-pi/2)
y-y1=m(x-x1)
y-3pi=6(x-pi/2)
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