Find the value?

Question

 

Let y=x*sqrt(1+x^2). When x=0 and dx=2, find the value of dy.

Answer: 2

Answer 1

y = x*sqrt(1+x^2) = x * (1+x^2)^(1/2)

 

dy = dx * sqrt(1+x^2) + x * [1/2 ((1+x^2)^(-1/2))*2*x*dx

 

When x=0 and dx=2,

 

dy = 2 * sqrt(1+0^2) + 0 * [1/2 ((1+0^2)^(-1/2))*2*0*2

 

dy = 2 * 1 + 0 / 1 = 2

Answer 2

At x = 0,

dy = y'dx = [sqrt(1+x^2) + 0] dx = dx = 2

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Attn:

y' = x' sqrt(1+x^2) + x[sqrt(1+x^2)]'  and

x[sqrt(1+x^2)]' = 0 at x = 0. (This step can be omitted since the second term contains a zero factor.)