2001 HSC - Q4(c) (1 Viewer)

[Stefano]

I feel sick. I'm having so much trouble with simple questions like this.
In the solutions they appear to be using the product rule. Why ?

I thought the solution to this was: 1/[1+(x/x+1)²] + 1/[1+(1/2x+1)²] (unsimplified)

 

[KFunk]

I feel sick. I'm having so much trouble with simple questions like this.
In the solutions they appear to be using the product rule. Why ?

I thought the solution to this was: 1/[1+(x/x+1)²] + 1/[1+(1/2x+1)²] (unsimplified)

Think of a general situation: let y = tan^-1[f(x)]. When you see this you do as substitution so:

y = tan^-1u ............................ u = f(x)

dy/du = 1/(1 + u²) .................du/dx = f'(x)

hence dy/dx = 1/(1 + f(x)²) . f'(x)

What you have forgotten to do is multiply by the derivative of the function within tan^-1.

1/[1+(x/x+1)²].d/dx(x/[x+1]) + 1/[1+(1/2x+1)²]d/dx(1/[2x+1]) should end up as zero.

 

[Stefano]

Think of a general situation: let y = tan^-1[f(x)]. When you see this you do as substitution so:

y = tan^-1u ............................ u = f(x)

dy/du = 1/(1 + u²) .................du/dx = f'(x)

hence dy/dx = 1/(1 + f(x)²) . f'(x)

What you have forgotten to do is multiply by the derivative of the function within tan^-1.

1/[1+(x/x+1)²].d/dx(x/[x+1]) + 1/[1+(1/2x+1)²]d/dx(1/[2x+1]) should end up as zero.

Aaahhhh!! Chain Rule! I didn't think it was necessary.

So, in summary: If y=tan^-1[F(x)], then:
dy/dx= f'(x)/{1+[f(x)]² } ?

 

[KFunk]

Yup, that's exactly it.

 

[Stefano]

Thank you once again KFunk! I owe you.