The Student Room Group

A simple differentiating problem

This is edexcel C4 differentiating question that i need help in. Thanks in advance!

Q: Find an expression in terms of x and y for dydx\frac{dy}{dx}, given that (xy)4=x+y+5(x-y)^4 = x+y+5

My attempt: 4dydx(xy)3=1+dydx\frac{-4dy}{dx}(x-y)^3 = 1+\frac{dy}{dx}

=> dydx+4dydx(xy)3=1\frac{dy}{dx}+\frac{4dy}{dx}(x-y)^3 = -1

Hence dydx=11+4(xy)3\frac{dy}{dx} = \frac{-1}{1+4(x-y)^3}

But according to the mark scheme of the textbook, the answer is 4(xy)311+4(xy)3\frac{4(x-y)^3-1}{1+4(x-y)^3}

Once again, help is appreciated!
Reply 1
Original post by Angry Citizen
This is edexcel C4 differentiating question that i need help in. Thanks in advance!

Q: Find an expression in terms of x and y for dydx\frac{dy}{dx}, given that (xy)4=x+y+5(x-y)^4 = x+y+5

My attempt: 4dydx(xy)3=1+dydx\frac{-4dy}{dx}(x-y)^3 = 1+\frac{dy}{dx}

=> dydx+4dydx(xy)3=1\frac{dy}{dx}+\frac{4dy}{dx}(x-y)^3 = -1

Hence dydx=11+4(xy)3\frac{dy}{dx} = \frac{-1}{1+4(x-y)^3}

But according to the mark scheme of the textbook, the answer is 4(xy)311+4(xy)3\frac{4(x-y)^3-1}{1+4(x-y)^3}

Once again, help is appreciated!


Your problem is with the first term

ddx((xy)4)\frac{d}{dx}((x-y)^4) requires the chain rule, as part of which you will need to find ddx(xy)\frac{d}{dx}(x-y)
Reply 2
ah i see now, thanks

Quick Reply

Latest