# Implicit Differentiation

# Implicit Differentiation

**Implicit Differentiation**

An **implicit relation** between x and y is one written as f(x,y)=g(x,y). They often appear for relations that it is impossible to write in the form y=f(x). Despite not having a nice expression for y in terms of x, we can still **differentiate** **implicit relations**.

**Method for Implicit Differentiation**

To carry out **implicit differentiation**, follow these steps.

**Step 1: Differentiate** terms that are in x only.

**Step 2: **Use the **chain rule** to **differentiate** terms in y only.

\dfrac{d}{dx}(f(y))=\dfrac{d}{dy}(f(y))\dfrac{dy}{dx}

This is the same as **differentiating** f(y) normally then **multiplying by** \dfrac{dy}{dx}.

**Step 3: **Use the **product rule** for terms that are in both x and y.

\dfrac{d}{dx}(u(x)v(y))=u(x)\dfrac{d}{dx}(v(y))+v(y)\dfrac{d}{dx}(u(x))

And you can use **the rule in step 2** to find \dfrac{d}{dx}(v(y)).

**Step 4: Rearrange the resulting equation** to make \dfrac{dy}{dx} the subject.

**Finding the Gradient with Implicit Differentiation**

**Implicit differentiation** still **finds the gradient**. We can calculate the **gradient** at a given x value by finding the y value from the **equation for the curve** then finding \dfrac{dy}{dx} at these values of x and y.

**We can also find other things**, such as an equation linking x and y at stationary points: \dfrac{dy}{dx}=0.

**Example: **A curve is defined by y^{2}-3xy+x^{2}=0. Find the **gradient** at all points where x=2.

**Implicitly differentiate**:

2y\dfrac{dy}{dx}-3x\dfrac{dy}{dx}-3y+2x=0

**Rearrange**:

(2y-3x)\dfrac{dy}{dx}-3y+2x=0

\dfrac{dy}{dx}+\dfrac{2x-3y}{2y-3x}=0

\dfrac{dy}{dx}=\dfrac{3y-2x}{2y-3x}

**Substitute** x value into curve to find y values:

At x=2, y^{2}-6y+4=0

y=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4\times1\times4}}{2\times1}

y=\dfrac{6\pm\sqrt{36-16}}{2}

y=3\pm\dfrac{\sqrt{20}}{2}

y=3\pm\sqrt{5}

**Substitute** x,y values into \dfrac{dy}{dx}:

y=3+\sqrt{5} \dfrac{dy}{dx}=\dfrac{3(3+\sqrt{5})-2\times2}{2(3+\sqrt{5})-3\times2} \dfrac{dy}{dx}=\dfrac{9+3\sqrt{5}-4}{6+2\sqrt{5}-6} \dfrac{dy}{dx}=\dfrac{5+3\sqrt{5}}{2\sqrt{5}} \dfrac{dy}{dx}=\dfrac{5\sqrt{5}+15}{10} \dfrac{dy}{dx}=\dfrac{3+\sqrt{5}}{2} |
y=3-\sqrt{5} \dfrac{dy}{dx}=\dfrac{3(3-\sqrt{5})-2\times2}{2(3-\sqrt{5})-3\times2} \dfrac{dy}{dx}=\dfrac{9-3\sqrt{5}-4}{6-2\sqrt{5}-6} \dfrac{dy}{dx}=\dfrac{5-3\sqrt{5}}{-2\sqrt{5}} \dfrac{dy}{dx}=\dfrac{5\sqrt{5}-15}{-10} \dfrac{dy}{dx}=\dfrac{3-\sqrt{5}}{2} |

Hence, **gradients** at x=2 are \dfrac{3+\sqrt{5}}{2} and \dfrac{3-\sqrt{5}}{2}

**Differentiating the Inverse Trigonometric Functions**

**Implicit differentiation** is how we find the **derivative** of \arcsin, \arccos and arctan.

**Example: **Find the **derivative** of \arcsin(x).

Take \sin of both sides:

\sin(y)=x**Differentiate (implicitly)**:

\cos(y)\dfrac{dy}{dx}=1

\dfrac{dy}{dx}=\dfrac{1}{\cos(y)}

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{\cos^{2}(y)}}

We know that \sin^{2}(y)+\cos^{2}(y)=1, so \cos^{2}(y)=1-\sin^{2}(y)

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-\sin^{2}(y)}}Now x=\sin(y) so we can **put this back into the equation**.

A Level

**Example 1: Implicit Differentiation**

4x^{3}+23\sin(x)\cos(y)+36xy^{2}+12y+1=0. Find \dfrac{dy}{dx}.

**[3 marks]**

**Differentiate implicitly**.

12x^{2}+23\cos(x)\cos(y)-23\sin(x)\sin(y)\dfrac{dy}{dx}+36y^{2}+72xy\dfrac{dy}{dx}+12\dfrac{dy}{dx}=0

12x^{2}+23\cos(x)\cos(y)+36y^{2}=23\sin(x)\sin(y)\dfrac{dy}{dx}-72xy\dfrac{dy}{dx}-12\dfrac{dy}{dx}

12x^{2}+23\cos(x)\cos(y)+36y^{2}=(23\sin(x)\sin(y)-72xy-12)\dfrac{dy}{dx}

\dfrac{dy}{dx}=\dfrac{12x^{2}+23\cos(x)\cos(y)+36y^{2}}{23\sin(x)\sin(y)-72xy-12}

A Level**Example 2: Inverse Trigonometric Functions**

Find the **derivative** of \arccos(x).

**[4 marks]**

Take \cos of both sides.

\cos(y)=x**Implicitly differentiate**.

-\sin(y)\dfrac{dy}{dx}=1

\dfrac{dy}{dx}=-\dfrac{1}{\sin(y)}

\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{\sin^{2}(y)}}

We know \sin^{2}(y)+\cos^{2}(y)=1, so \sin^{2}(y)=1-\cos^{2}(y)

\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-\cos^{2}(y)}}Now x=\cos(y) so we can **put this back into the equation**.

A Level

## Example Questions

**Question 1: **A curve is 9\ln(x)\cos(y)=0. Find \dfrac{dy}{dx}.

**[2 marks]**

Differentiate implicitly:

\dfrac{9\cos(y)}{x}-9\ln(x)\sin(y)\dfrac{dy}{dx}=0

\dfrac{9\cos(y)}{x}=9\ln(x)\sin(y)\dfrac{dy}{dx}=0

\dfrac{dy}{dx}=\dfrac{9\cos(y)}{9x\ln(x)\sin(y)}

\dfrac{dy}{dx}=\dfrac{\cos(y)}{x\ln(x)\sin(y)}

**Question 2: **Find the gradient of the curve 3x^{2}y-x=0 when x=\dfrac{1}{3}

**[3 marks]**

Differentiate implicitly.

6xy+3x^{2}\dfrac{dy}{dx}-1=0

3x^{2}\dfrac{dy}{dx}=1-6xy

\dfrac{dy}{dx}=\dfrac{1-6xy}{3x^{2}}

Find y when x=\dfrac{1}{3}.

3\left(\dfrac{1}{3}\right)^{2}y-\dfrac{1}{3}=0

3\times\dfrac{1}{9}y-\dfrac{1}{3}=0

\dfrac{1}{3}y-\dfrac{1}{3}=0

\dfrac{1}{3}y=\dfrac{1}{3}

y=1

Substitute into our expression for \dfrac{dy}{dx}:

\dfrac{dy}{dx}=\dfrac{1-6\times1\times\dfrac{1}{3}}{3\left(\dfrac{1}{3}\right)^{2}}

\dfrac{dy}{dx}=\dfrac{1-2}{3\times\dfrac{1}{9}}

\dfrac{dy}{dx}=\dfrac{-1}{\dfrac{1}{3}}

\dfrac{dy}{dx}=-3

So the gradient is -3 at this point.

**Question 3:**

a) Find the implicit derivative of 3x^{3}y^{2}-xy=1

b) Find an equation linking x and y at the stationary points of the curve.

c) Use this equation and the equation of the curve to find the stationary points of the curve.

**[8 marks]**

a) 3x^{3}y^{2}-xy=1

9x^{2}y^{2}+6x^{3}y\dfrac{dy}{dx}-y-x\dfrac{dy}{dx}=0

9x^{2}y^{2}-y=x\dfrac{dy}{dx}-6x^{3}y\dfrac{dy}{dx}

9x^{2}y^{2}-y=(x-6x^{3}y)\dfrac{dy}{dx}

\dfrac{dy}{dx}=\dfrac{9x^{2}y^{2}-y}{x-6x^{3}y}

b) Stationary point is at \dfrac{dy}{dx}=0

0=\dfrac{9x^{2}y^{2}-y}{x-6x^{3}y}

9x^{2}y^{2}-y=0

y(9x^{2}y-1)=0

y=0 or 9x^{2}y-1=0

y=0 or 9x^{2}y=1

c) Curve equation is 3x^{3}y^{2}-xy=1

y=0 gives 0=1 so we cannot use this result.

9x^{2}y=1

y=\dfrac{1}{9x^{2}}

Substitute this into the curve.

3x^{3}\left(\dfrac{1}{9x^{2}}\right)^{2}-\dfrac{x}{9x^{2}}=1

\dfrac{3x^{3}}{81x^{4}}-\dfrac{1}{9x}=1

\dfrac{1}{27x}-\dfrac{1}{9x}=1

1-3=27x

x=-\dfrac{2}{27}

y=\dfrac{1}{9x^{2}}

y=\dfrac{1}{9\left(-\dfrac{2}{27}\right)^{2}}

y=\dfrac{1}{9\left(\dfrac{4}{729}\right)}

y=\dfrac{1}{\left( \dfrac{4}{81}\right) }

y=\dfrac{81}{4}

So the stationary point is \left(-\dfrac{2}{27},\dfrac{81}{4}\right)

**Question 4: **Find the derivative of y=\arctan(x).

**[3 marks]**

Take \tan of both sides.

\tan(y)=x

Differentiate implicitly:

\sec^{2}(y)\dfrac{dy}{dx}=1

\dfrac{dy}{dx}=\dfrac{1}{\sec^{2}(y)}

Identity: \sec^{2}(y)=1+\tan^{2}(y)

\dfrac{dy}{dx}=\dfrac{1}{1+\tan^{2}(y)}

Put x=\tan(y) back into the equation.

\dfrac{dy}{dx}=\dfrac{1}{1+x^{2}}

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