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Derivative of x=function of y

Hi

Bit stuck on this bit:

so x=(y/6)^2 + 1/3

Find an equation of tangent to the curve at point where y=2

so dy/dx = 3/y

y = 2
so x = 1

tangent=3/y

However im a bit confused since it is expressed as x instead of y
so would i just use y-y1=m(x-x1)?

Any help would be great
Reply 1
Avatar for Swayum
Swayum
18
You've differentiated incorrectly.

x = (y/6)^2 + 1/3 = y^2/36 + 1/3

But then, yeah, just plug in y = 2 into dy/dx and use the formula.
Reply 2
Avatar for bigbosss
bigbosss
hey use the formula rearrange to get y and you should get y=3/2x+1/2, hope this helps!

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