How do you find the tangent line of #f(x) = 3x/sinx^2 # at x=5?

1 Answer
Oct 24, 2017

#y = -754.2x + 3806#

Explanation:

Use the quotient rule when your function can be thought of as one function divided by another.

if #f(x) = (g(x))/(h(x))#
then #f'(x) = (g'(x)h(x)-g(x)h'(x))/(h^2(x))#

so if #f(x) = (3x)/sinx^2#
then #f'(x) = (3sinx^2 - 3x(2xcosx^2))/sin^2x^2#

so #f'(x) = (3sinx^2 - 6x^2cosx^2)/sin^2x^2#

Now I'm gonna assume we're working in degrees because #x=5# seems more like a value for degrees than radians, but I could be wrong. If I am the steps still apply it's just that my numbers will be off.

The derivative will give us the gradient, #m#, of the tangent, #y=mx+c#, at any given point #x#.

#m = (3sin25-150cos25)/(sin25)^2 = -134.7/0.1786#

#m = -754.2#

If you work with radians you should get around #-8520# as your gradient or something.

Now we have the equation

#y = -754.2x + c#

for which we still need the value for #c#, which we can obtain by substituting a point #(x,y)# which is on the tangent line.

We are already given a point #x=5#, so we should work out the corresponding #y# value:

#y = (3x)/(sinx^2) =15/sin25 = 35.49#

#:.#

#(x,y)=(5.000, 35.49)#

which we can substitute in to give us

#35.49 = -754.2 xx 5.000 + c#

#35.49 + 3771 = 3806 = c#

so

#y = -754.2x + 3806#

Obviously this will be a very different equation using radians, but the general principles are the same. I've also used all values to 4 significant figures.