Calculus: Chain Rule and Tangent Equations?

Question

The Question is: Find the tangent equation when x = 0 for y = [x+ (x+x^1/2)^1/2]^1/2. So, basically i just need to know how to find the derivative of this function with the chain rule so i can get the slope of the tangent line.

Thanks for all the help guys!

Answer 1

Do you mean y = 0? Well, why not? It might be that y=0 is the tangent line..

At first sight it looks as if the derivative is undefined at x = 0, because y is undefined for x<0.

You will only be able to find the right hand derivative NEAR x = 0. Then you might be able to see what happens when x-> 0 through positive values.

You could use logarithmic derivative for the first stage, but the chain rule is heavily involved.

Answer 2

The derivative is

y' = (1/2) [x+ (x+x^1/2)^1/2]^(-1/2) * [1 + (1/2) (x+x^1/2)^(-1/2)] * [1 + (1/2) x^(-1/2)], for x >0.

This function is not diffrenetible at x = 0, the tangent is vertical.

Answer 3

It is undefined because there is a division by 0 in the answer (there's too much working to show here but there's an x^-1/2 in it), so this could mean that the gradient is infinite, and when you check it using a program like autograph or omnigraph, it is!

Answer 4

y = f(x)
f(0) = 0
f'(0) = infinity
Equation of the tangent line: x = 0
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Ideas to save time: When you take derivative at x = 0, you always have a non-zero number divided by zero. Therefore, f'(0) = infinity.

Answer 5

Wow! What a mess:

dy/dx = 1/2[x + (x + x^1/2)^1/2]^-1/2 multipled by

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

I don't think I will try to simplify this monster.

Answer 6

slope will be undefined as denominator will be zero.

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

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

=>1/2[x +sqrt(x+sqrt(x)]*d[[x +(x + x^(1/2))^1/2]

now evaluate d[[x +(x + x^(1/2))^1/2]

d[[x +(x + x^(1/2))^1/2] =

(1 +(1/2)(x+x^1/2)^(-1/2)*d(x+x^1/2)

=>(1 +(1/2)(x+x^1/2)^(-1/2)*(1 + 1/2(x^(-1/2))

=>(1 + (1/2sqrt(x+sqrt(x))*(1 +1/2sqrt(x))

=>[(2sqrt((x+sqrt(x))+1)/(2sqrt((x+sqrt(x))]*2sqrt(x)+1)/2sqrt(x)

=>[(sqrt(x + sqrt(x)) + 1]/sqrt((x+sqrt(x))*[(sqrt(x)+1)/sqrt(x)

=>[(sqrt(x + sqrt(x))) + 1]*[(sqrt(x)+1)]/[(sqrt(x+sqrt(x))*sqrt(x)]

y' =

[(sqrt(x + sqrt(x))) + 1]*[(sqrt(x)+1)]
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2(x +sqrt(x+sqrt(x)))]* [(sqrt(x+sqrt(x))*sqrt(x)]