Implicit differentiation

Differentiation: Implicit KUS objectives BAT Understand Implicit differentiation and find the gradient function for functions in x and y Starter differentiate: 𝑒 3𝑥 + 𝑥 2 2 𝑥 3 +3 2 𝑥 sin 𝑥 𝑒 2𝑥 2 𝑥 2 −5 4 𝑥 3 tan 𝑥 4𝑦(1+3 𝑥 2 ) !!!!!

Notes Implicit differentiation For some equations it is impossible to rearrange to give y = f(x) and hence differentiate. One approach is to use the chain rule to differentiate each term without rearrangement For example differentiate y2 Think Pair share Now we can do Not possible to separate the variables – do by inspection

Notes Implicit differentiation Implicit effectively means all the terms are mixed up, not necessarily in the form ‘y = …’ This technique is useful as some equations are difficult to arrange into this form…     Write dy/dx after differentiating the y term This is written differently  The reason is that as the equation is not written as ‘y =…’, y is not a function of x Differentiate the y term as you would for an x term   For example:       This is what happens when you differentiate an equation which starts ‘y =…’    

One approach is to use the chain rule to differentiate each term without rearrangement Key examples: 𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 function Gradient function 𝑦 4   1 𝑦 𝑦 sin 𝑦 𝑥 2 + 𝑦 2 𝑒 𝑦 ln 𝑦 4 𝑦 3 𝑑𝑦 𝑑𝑥 1 𝑦 2 𝑑𝑦 𝑑𝑥 1 2 𝑦 𝑑𝑦 𝑑𝑥 cos 𝑦 𝑑𝑦 𝑑𝑥 2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 𝑒 𝑦 𝑑𝑦 𝑑𝑥 1 𝑦 𝑑𝑦 𝑑𝑥

Find dy/dx in terms of x and y for the following equation: WB 1 Find dy/dx in terms of x and y for the following equation: 𝑥 3 +𝑥+ 𝑦 3 +3𝑦=6   It would be very difficult to rearrange this into the form ‘y = …’ Differentiate each part one at a time                   Isolate the terms with dy/dx in               Factorise by taking out dy/dx             Divide by (3y2 + 3)       You now have a formula for the gradient, but in terms of x AND y, not just x

WB 2 Derivative of product xy 𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 Key example: differentiate f(x, y) = xy using the Product rule Now we can do

Practice 1 finding gradient function 𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 Work out the derivatives of each of these: Since x and y are > 0 no modulus signs needed

WB 3 Find the value of dy/dx at the point (1,1) where: 4𝑥 𝑦 2 −5𝑥=10 𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥   Differentiate each one at a time – remember the product rule for the first term!!!!                         Add 5, subtract 4y2   You can substitute x = 1 and y = 1 now (or rearrange first)  Do not replace the terms in dy/dx   Divide by 8   Keep your workings tidy – you may need to use the product rule on multiple terms as well as factorise – show everything you’re doing!

rearranges to 𝑑𝑦 𝑑𝑥 =− 𝑥 𝑦 WB 4 A circle has equation 𝑥 2 + 𝑦 2 =25 Find two equations of tangents to the curve when x = 4 𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 differentiating 2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 =0 rearranges to 𝑑𝑦 𝑑𝑥 =− 𝑥 𝑦 when 𝑥=4 𝑦 2 =25− 4 2 𝑦 =±3 Since x and y are > 0 no modulus signs needed At (4, 3) 𝑑𝑦 𝑑𝑥 =− 4 3 At (4, -3) 𝑑𝑦 𝑑𝑥 = 4 3 y−3=− 4 3 (𝑥−4) y+3= 4 3 (𝑥−4) Tangent 4𝑥+3𝑦=25 Tangent 4𝑥−3𝑦=25

Practice 2 finding gradient function Find an expression in x and y for the gradient function of each of these:

M1A1 M1A1 A1 normal M1A1 equation WB 5 A curve C is described by the equation 3 𝑥 2 −4 𝑦 2 +4𝑥−5𝑦+12=0 Find an equation of the normal to C at point (3, 3), giving your answer in the form 𝑎𝑥+𝑏𝑦+𝑐=0 M1A1 M1A1 A1 normal equation M1A1

WB 6 Find the value of dy/dx at the point (1,1) where 𝑒 2𝑥 ln 𝑦 =𝑥+𝑦−2 Give your answer in terms of e         Differentiate (again watch out for the product rule!               Sub in x = 1 and y = 1       ln1 = 0 since e0 = 1, cancelling the term out     Subtract dy/dx   Factorise   Divide by (e2 – 1)  

Write one thing you have learned Write one thing you need to improve KUS objectives BAT Understand Implicit differentiation and find gradient function for functions in x and y Crucial points Make sure that you understand the process of differentiating an equation implicitly   Write one thing you have learned Write one thing you need to improve

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