Differentiating a Modulus Function

If you differentiate a modulus function, does the modulus bit dissapear?

I just ask because an AQA practice paper suggests so in terms of the answer it gives.

Question:
Find the angle between the tangents to the curves (y=x^2 - 4x & y=abs(4x - x^2)) at the point (4, 0).

Answer given: 151.9deg

However, this assumes the modulus bit dissapears when you differentiate to find the gradient of the original modulus function.
This would make the gradient of the modulus function -4 at point (4,0). However, looking at the graphs of these equations, it would appear that the gradient runs along the line y=0.

Very, very confused.
Please help.

Reply 1

meef cheese

10

Steezy20

If you differentiate a modulus function, does the modulus bit dissapear?

I just ask because an AQA practice paper suggests so in terms of the answer it gives.

Question:
Find the angle between the tangents to the curves (y=x^2 - 4x & y=abs(4x - x^2)) at the point (4, 0).

Answer given: 151.9deg

However, this assumes the modulus bit dissapears when you differentiate to find the gradient of the original modulus function.
This would make the gradient of the modulus function -4 at point (4,0). However, looking at the graphs of these equations, it would appear that the gradient runs along the line y=0.

Very, very confused.
Please help.

If u square both sides, the modulus disappears and u can use implicit differentiation

e.g y=|x|
=> y² = x²
=>2y.dy/dx = 2x
=>dy/dx = x/y
=> dy/dx = x/|x|

Reply 2

Steezy20

OP

1

Cheers.

However, I'm presuming there's another way given that implicit differentiation is only introduced in the module after this practice paper (I only began to learn implicit differentiation a week ago).

Just seems silly that AQA would use a question that can only be answered if you've studied further.

Cheers anyway.

Reply 3

RichE

15

I'm pretty uncomfortable with the question as

y = |x(x-4)| isn't differentiable at x=4.

There isn't a well-defined single tangent to the curve there.

Reply 4

Steezy20

OP

1

RichE

I'm pretty uncomfortable with the question as

y = |x(x-4)| isn't differentiable at x=4.

There isn't a well-defined single tangent to the curve there.

Thats exactly what I thought.

Reply 5

SsEe

13

Um. It's a badly asked question. But I guess they mean

Find the angle between the tangents to the curves y=x^2 - 4x & y=4x - x^2 at the point (4, 0).
So gradients are 4 and -4 repectively. arctan(4) - arctan(-4) = 151.9deg

Differentiating a Modulus Function

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