Query in Implicit Differentiation (Regarding chain rule)

Question

We all are familiar with the ladder problem in implicit differentiation,

Here the ladder is 10 ft long and sliding from the wall at rate of 1 ft/sec. i.e dy/dx =1 ft/sec.

I had to find out how fast the top of ladder is sliding down when the bottom is 6 ft away from wall. I.E I HAD to find out dy/dx.

We know that the figure(plz draw) makes a right angled triangle with hypotenuse known and two legs unknown.

we relate these variables using pythagoras theorem.

x² + y² = 10² i.e x² + y² = 100

we now differentiate this equation w.r.t time

2x dx/dt + 2y dy/dt = 0

we get, dy/dt = -3/4 ft/sec.

i have no issues here what i want to know is

the first step, when we start differentiating, 2x dx/dt+2y dy/dt.

Now here we apply chain rule i know that but when we differentiate x² with respect to x, we get 2x.

Is it like f(x)= x², i mean what is x² here is it like y=x²,and we perform dy/dx. i know i am wrong but can someone explain that. also incase of 2y i.e y²

Answer 1

When you take d/dx of (x²), you get 2x dx/dx or 2x

When x and y are functions of t only (x and y are tied together through the constraint equation (PT)), then

d/dt of x² by chain rule will be df(x)/dx dx/dt or 2x dx/dt which is how we get the rate relationship.

d/dt of y² is the same: dg(y)/dy dy/dt or 2y dy/dt

I hope that helped.

You can also have a complex relationship between y(x) and x and take the implicit derivative of y and a standard derivative of x

sqrt(y) sinx = 3 By Imp Diff: 1/2 * (sinx/sqrt(y)) dydx + sqrt(y) cos(x) = 0 and you can solve for dy/dx

(Yes, it may be easier to just solve y as f(x) and take the derivative.)