How to find gradient of tangent to circle x^2+y^2=1 at given point using implicit differentiation

Differentiating a function or relation where y cannot be made the subject of the formula explicitly requires a method of differentiation called "implicit differentiation".

The theory behind it is quite simple and is based the idea that d/dx (y) is dy/dx or y'. If y is part of a composite function, a product or a quotient then the chain, product or quotient rules may be applied to differentiate terms containing y.

In this example, we are asked for the gradient of the tangent to the circle x^2 + y^2 = 1 at the point (cos 70°, sin 70°), correct to the 3 significant figures.

Technically, y cannot be made the subject of the formula explicitly without considering just the positive semicircle. The equation can be differentiated implicitly with ease without the need for taking the square root of 1 - x^2.

Differentiating implicitly is done by differentiating each term of the equation of the circle with respect to x:

d/dx (x^2) + d/dx (y^2) = d/dx (1) which gives 2x + 2y × y' = 0

Making y' the subject of the formula we obtain y' = -x / y or dy/dx = -x / y.

Substituting in the point of interest into the derivative to the find the gradient of the tangent:

y' = -cos 70° / sin 70° which gives -0.364 (correct to 3 significant figures).