# Find the derivative

Find the derivative;

G(v)= (v^3-1)/(v^3+1)

Find the limit;

lim(sin3x)/(sin5x)

x->0

Find the derivative;

R(w)= (cosw)/(1-sinw)

H(o)=(1+seco)/(1-seco)

Find the derivative;

F(x)= cos(3x^2)+{cos^2}3x

N(x)=(sin5x-cos5x)^5

"Assume that the equation determines a differentiable function f such that y=f(x), find y'."

5x^2-xy-4y^2=0

x^4+4x^2y^2-3xy^2+2x=0

"Assume all variables are functions of t."

If S=z^3 and dz/dt=-2 when z=3, find dS/dt.

Find the extrema of f on the given interval.

f(x)=3x^2-10x+7; [-1,3]

"Show that f satisfies the hypotheses of Rolle's Theorem on [a,b] and find all numbers c in (a,b) such that f'(c)=0."

f(x)=5-12x-2x^2; [-7,1]

"Determine whether f satisfies the hypotheses of the mean value theorem on [a,b] and if so, find all numbers c in (a,b) such that f(b)-f(a)=f'(c)(b-a)."

f(x)=3x^2+x-4; [1,5]

Find the extrema and sketch the graph of f.

f(x)=(2x-5)/(x+3)

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Calculus problems

Some of the questions I can't understand what they are asking. Those are in quotes. I also need all the calculations for all the problems so I can see how you got the answers. Thanks!

Find the derivative;

Using the quotient rule, we have

Find the limit;

as the value of each of the limits involved is equal to 1.

Find the derivative;

Using the quotient rule, we have

Find the derivative

Find the derivative;

Here we use the chain rule or function of a function rule to differentiate.

Again, we use the chain rule here.

"Assume that the equation determines a differentiable function f such that y=f(x), find y'."

The following functions are called implicit functions. It is difficult to solve for one of the variables involved to be expressed in terms of the other variable, in which case the differentiation is difficult. The above statement is to assume that the y resulted from solving this equation can be differentiated so that we can find . This process of differentiation is called 'Implicit differentiation'. In this process, any term that involves y is considered as function of y and y is considered as function of x.

Differentiating with respect to x, we get

Solving for , gives

Differentiating with respect to x, we get

Solving for , gives

"Assume all variables are functions of t."

In a situation when one variable is given in terms of another variable, they are actually connected to another independent variable, we carry out this kind of differentiation.

If and when z=3, find .

Differentiating both sides with respect to t, we get .

Given when z=3 and , we have

Find the extrema of f on the given interval.

; [-1,3]

Differentiating with respect to x, we get . For extreme values, we equate the derivative to zero. This gives . Further differentiating, we get for . The function has local minimum at .

As the function is given in a closed interval, we compare the values . We have .

Hence the maximum value is attained at x=-1 and local minimum at

"Show that f satisfies the hypotheses of Rolle's Theorem on [a,b] and find all numbers c in (a,b) such that f'(c)=0."

The Hypothesis of Rolle's theorem is as follows

If f(x) is continuous on [a, b], differentiable on (a, b) and f(a)=f(b), then there exists an element c in (a, b) such that f'(c)=0

; [-7,1]

Now, we have a=-7 and b=1. As the given function is a quadratic polynomial, the first two statements of Rolle's theorem are true for this function.

Now, f(-7)=5+84-98=-9 and f(1)=1-12-2=-9. Hence the condition f(a)=f(b) is also satisfied. Now, f'(x)=0 gives -12-4x=0 which implies x=-3. The point where the derivative (ie c in the Rolle's Theorem) vanishes is c=-3 which can be easily verified that it lies in (-7, 1)

"Determine whether f satisfies the hypotheses of the mean value theorem on [a,b] and if so, find all numbers c in (a,b) such that f(b)-f(a)=f'(c)(b-a)."

The Hypothesis of Mean value theorem is as follows

If f(x) is continuous on [a, b], differentiable on (a, b), then there exists an element c in (a, b) such that f(b)-f(a)=f'(c)(b-a) ie

; [1,5]

Given function is a polynomial function, hence the conditions of Mean Value theorem are satisfied. Here a=1, b=5

Differentiating, we have f'(x)=6x+1

We compute . Equating this value to the derivative, we have 6x+1=19 which implies x=3 and it lies in (1, 5).

Hence the value of c we are looking for is c=3.

Find the extrema and sketch the graph of f.

Differentiating, we get

This implies the function has positive derivative for all real values of x except -3 where it is undefined.

Hence the function will have extrema only if the domain is taken as closed interval. As we are not given any such interval, this function has no extrema.

Sketching the graph of this function, we have

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