To answer problem no. 1 we have to recall the formula for chain rule .

The derivative of a function like (c(x))^n is n*(c(x)^(n-1)) * c'(x) .

With this, we can get the derivative of F(x) = f(x^5) by taking the derivative of the outer function f first then using...

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To answer problem no. 1 we have to recall the formula for chain rule .

The derivative of a function like (c(x))^n is n*(c(x)^(n-1)) * c'(x) .

With this, we can get the derivative of F(x) = f(x^5) by taking the derivative of the outer function f first then using the chain rule for the inner function x^5, so F'(x) = f'(x^5)*5x^(5-1) simplifying this we will have F'(x) = 5f'(x^5)*x^4.

To get F'(a) we plug-in a for x in the derivative we got, so F'(a) = 5f'(a^5)*(a^4) and using the information given in the problem it will be 5*(12)*(13)=780.

The same with F'(a), we get G'(a) using the chain rule. However, in this case it is in the outer function that applies. G(x) = (f(x))^5 has the derivative of G'(x) = 5(f(x))^(4)*f'(x).

So, G'(a)= 5(f(a))^(4)*f'(a).

Again, referring to the given information we have G'(a)=5*(3)^(4)*9 = 3645.

(2) We are given that the rate at which water is leaving the tank is ` `` ``(dV)/(dt)=R-9500"cc"/"min" ` where R is the unknown rate of additional water being poured into the tank.. The height H of the tank is 6m, the diameter of the top is 6.5m. The rate at which the depth of the water is increasing is `(dh)/(dt)=24"cm"/"min" ` where h is the depth of the water when h=3.5m.

The volume of a right circular cone is `V=(pi r^2h)/3 ` . ` `If the depth of the water is h and the radius of the cone containing the water is r, then by similar triangles we have `3.25/6=r/h ` . (3.25m is the radius of the circular opening of the tank, and 6m is the height of the tank.) Then we can express r as a function of h: `r=3.25/6*h ` .

Substituting we get:

`V=1/3*pi*(3.25/6h)^2*h ` or `V=(169pih^3)/1728 `

Then `(dV)/(dt)=(169pih^2)/576 (dh)/(dt) `

Now with h=3.5m or 350cm, ` (dh)/(dt)=24"cm"/"min" ` ` `

`R-9500"cc"/"min"=(169pi(350"cm")^2)/576*24"cm"/"min" `

`R-9500=(496860000pi)/576 "cc"/"min" `

`R~~2719450.9"cc"/"min" ` or `R~~2.72"m"^3/"min" `

(3) We know ` A=pir^2 ` so `(dA)/(dt)=2pir(dr)/(dt) ` .

Here we are given `(dr)/(dt)=2,r=2 ` so `(dA)/(dt)=2pi(2)(2)=8pi~~25.1 `

(4) Given `xy=2,(dy)/(dt)=3, x=2 ` :

We have `x(dy)/(dt)+y(dx)/(dt)=0 ` Substituting the known values:

`2(3)+1(dx)/(dt)=0 ` (xy=2 implies y=1 when x=2.)

`(dx)/(dt)=-6 `

(5) The volume of a sphere is `V=4/3pir^3 ` or in terms of the diameter

`V=4/3pi(d/2)^3=(4pid^3)/24=(pid^3)/6 `

Then `(dV)/(dt)=(pid^2)/2 (dd)/(dt) ` Substituting known values we get:

`(dV)/(dt)=(pi*100)/2(.1)=5pi~~15.71("cm"^3)/"min" `

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