Question

We know W = F×S, let this be the equation 1; Also dW = F×dS
But if we differentiate equation 1,
by the product-rule dW = d(F×S) = F×dS+S×dF
So where does this S×dF term go?

Answer 1

Answer:

The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that f is differentiable at the point P(x0,y0), where x0=g(t0) and y0=h(t0) for a fixed value of t0. We wish to prove that z=f(x(t),y(t)) is differentiable at t=t0 and that Equation 4.29 holds at that point as well.

Since f is differentiable at P, we know that

z(t)=f(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)+E(x,y),

4.30

where lim(x,y)→(x0,y0)

E(x,y)

√

(x−x0)2+(y−y0)2

=0. We then subtract z0=f(x0,y0) from both sides of this equation:

Answer 2

$\huge\sf\red{We know W = F×S, let this be the equation 1; Also dW = F×dS</p><p>But if we differentiate equation 1,by the product-rule dW = d(F×S) =F×dS+S×dFSo where does this S×dF term go?}$

Answer 3

$\huge\sf\red{We know W = F×S, let this be the equation 1; Also dW = F×dS</p><p>But if we differentiate equation 1,by the product-rule dW = d(F×S) =F×dS+S×dFSo where does this S×dF term go?}$