**Differentiation – Implicit Function**

i) In given implicit function, we solve this for dy/dx directly

ii) Moderate level sometime in question, we asked dy/ dx at some particular point x.

So, after getting dy/dx, we put the given particular value of x in dy/ dx and get the desired result.

If the relation between the variable x and y is given by an equation containing both of the form φ (x, y) = 0 and this equation is not immediately solvable for y i, e., it is not passible to express y as a function of x in the form y = φ(x), then y is called an implicit function of x. In such a case to find dy/dx differentiate both sides of the given relation with respect to ‘x’ keeping in the mind that the derivative of f(y) with respect ‘x’ is (dφ/dy). (dy/dx).

**Examples:**

- \(\frac{d}{dx}(\tan y)={{\sec }^{2}}y.\frac{dy}{dx}\).
- \(\frac{d}{dx}({{y}^{4}})=4{{y}^{3.}}.\frac{dy}{dx}\).

**Shortcuts for implicit function: **For implicit function, put

\(\frac{d}{dx}\{f(x,y)\}=\frac{-\partial f/\partial x}{\partial f/\partial y}\),

Where \(\frac{\partial f}{\partial x}\) is a partial differential of given function with respect to ‘x’ and \(\frac{\partial f}{\partial y}\) means partial differential of given function with respect to ‘y’.

**Problem: **If sin²y + cos (xy) = k, then dy/dx =?

**Solution:** Given that

sin²y + cos (xy) = k,

On differentiation both sides with respect to ‘x’ we get

\(\frac{d}{dx}\left( {{\sin }^{2}}x+\cos xy \right)=\frac{d}{dx}(k)\).

\(\frac{d}{dx}({{\sin }^{2}}x)+\frac{d}{dx}(\cos xy)=0\).

\(2\sin y\cos y.\frac{dy}{dx}+(-\sin xy)\frac{d}{dx}(xy)=0\).

\(\sin 2y\frac{dy}{dx}-\sin xy\left( x\frac{dy}{dx}+y.1 \right)=0\).

\(\sin 2y.\frac{dy}{dx}-x\sin xy.\frac{dy}{dx}=y\sin xy\).

\(\frac{dy}{dx}=\frac{y\sin (xy)}{\sin 2y-x\sin (xy)}\).