This question has been giving me some problems...Please help.

[CHiMER4]

Given that y=sqrt(x/(x+1))

the gradient of the normal to the curve at any point, is given by the formula -1[sqrt(px(x+1)^q)]

Determine the values of p and q where p, q ∈ Q

my approach:

y'=-1/[2(sqrt(x/(x+1)))]

is my derivative correct?

If so, how should I proceed?

 

[Deleted member 4993]

Given that y=sqrt(x/(x+1))

the gradient of the normal to the curve at any point, is given by the formula -1[sqrt(px(x+1)^q)]

Determine the values of p and q where p, q ∈ Q

my approach:

y'=-1/[2(sqrt(x/(x+1)))]

is my derivative correct?

If so, how should I proceed?

is my derivative correct?

No! You have to use Chain Rule.

If I were to do do this problem, I would rewrite the function as:

y = \(\displaystyle \sqrt{ 1 - \frac{1}{x+1}}\)....... not absolutely necessary but in my eyes it looks less complicated

y = [ 1 - (x+1)^-1]^1/2

y' = (1/2) * [ 1 - (x+1)^-1]^(-1/2) * (x+1)^-2

 

[apple2357]

Remember the derivative gives you the gradient of the tangent. Do you know how to get the gradient of the normal from the tangent?
I haven't checked your y'. Can you share your method?

 

[BigBeachBanana]

Remember the derivative gives you the gradient of the tangent. Do you know how to get the gradient of the normal from the tangent?
I haven't checked your y'. Can you share your method?

The normal curve is perpendicular to the tangent curve (if I interpreted your question correctly).

 

[CHiMER4]

is my derivative correct?

No! You have to use Chain Rule.

If I were to do do this problem, I would rewrite the function as:

y = \(\displaystyle \sqrt{ 1 - \frac{1}{x+1}}\)....... not absolutely necessary but in my eyes it looks less complicated

y = [ 1 - (x+1)^-1]^1/2

y' = (1/2) * [ 1 - (x+1)^-1]^(-1/2) * (x+1)^-2

Thank you.

 

[CHiMER4]

Remember the derivative gives you the gradient of the tangent. Do you know how to get the gradient of the normal from the tangent?
I haven't checked your y'. Can you share your method?

Since tangent and normal are perpendicular , it means that mTangent/y'=mNormal.
Therefore ...Is the working below correct or have I gone wrong somewhere?