# Calculus Problems Solved Step By Step

Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.Jump down this page to: [Power rule, $x^n$] [Exponential, $e^x$] [Trig derivatives] [Product rule] [Quotient rule] [Chain rule]$$\frac\text = 0 \quad \frac \left(x\right) = 1$$ $$\frac \left(x^n\right) = nx^$$\begin \frac\left( e^x \right) &= e^x &&& \frac\left( a^x \right) &= a^x \ln a \ \ \end\begin \frac\left(\sin x\right) &= \cos x &&& \frac\left(\csc x\right) &= -\csc x \cot x \ \ \dfrac\left(\cos x\right) &= -\sin x &&& \frac\left(\sec x\right) &= \sec x \tan x \ \ \dfrac\left(\tan x\right) &= \sec^2 x &&& \frac\left(\cot x\right) &= -\csc^2 x \end Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.Recall that $x^a x^b = x^.$ $\begin \sqrt\left(x^2 – 8 \frac \right) &= x^x^2 – 8x^ x^x^ \[8px] &= x^\, – 8x^\, x^ \end$ We can now take the derivative: $\begin \dfrac \left(\sqrt\left(x^2 – 8 \frac \right) \right) &= \dfrac\left(x^\right) – 8\dfrac\left(x^ \right) \dfrac \left( x^ \right) \[8px] &= \frac x^ – 8 \left(\frac x^ \right) \left(-\frac \right)x^ \[8px] &= \fracx^\, – 4 x^ \,- \frac x^ \quad \cmark \end$ To use only the Power Rule to find this derivative, we must start by expanding the function so we can proceed term by term: $\begin \left(2x^2 1 \right)^2 &= (2x^2)^2 2(2x^2)(1) 1^2 \[8px] &= 4x^4 4x^2 1 \end$ We can now take the derivative: $\begin \dfrac\left(2x^2 1 \right)^2 &= \dfrac\left(4x^4 \right) \dfrac \left(4x^2 \right) \cancelto \[8px] &= 4(4)x^3 4(2)x \[8px] &= 16x^3 8x \quad \cmark \end$ Recall that $a^ = a^n a^m$.Hence $$e^ = e \cdot e^x$$ And remember that $e$ is just a number — a constant.

For now, to use only the Power Rule we must multiply out the terms.Hence, you already know how to do all of the following steps; the only new part to maximization problems is what we did in Stage I above. We want to minimize the function $$A(r) = 2\pi r^2 \frac$$ and so of course we must take the derivative, and then find the critical points. $\begin A'(r) &= \dfrac\left(2\pi r^2 \frac \right) \[8px] &= \dfrac\left(2\pi r^2 \right) \dfrac\left(\frac \right) \[8px] &= 2\pi \dfrac\left(r^2 \right) 2V \dfrac\left(r^ \right) \[8px] &= 2\pi(2r) 2V \left((-1)r^ \right)\[8px] &= 4 \pi r\, – \frac \end$The critical points occur when $A'(r) = 0$: $\begin A'(r) = 0 &= 4 \pi r\, – \frac \[8px] \frac &= 4 \pi r \[8px] \frac &= r^3 \[8px] r^3 &= \frac \[8px] r &= \sqrt[3] \end$ We thus have only one critical point to examine, at $r = \sqrt[3]\,.$Step 5.Next we must justify that the critical point we’ve found represents a minimum for the can’s surface area (as opposed to a maximum, or a saddle point).Recall that $\dfrac\left(x^n\right) = nx^.$ $\begin \dfrac\left( \fracx^9\right) &= \frac \dfrac\left(x^9 \right) \[8px] &= \frac\left(9 x^ \right) \[8px] &= \frac(9) \left(x^8 \right) \[8px] &= 6x^8 \quad \cmark \end$ Recall that $\dfrac\left(x^n\right) = nx^.$ We simply go term by term: $\begin \dfrac\left(2x^3 – 4x^2 x -33 \right) &= \dfrac\left( 2x^3\right) – \dfrac\left( 4x^2\right) \dfrac(x) – \cancelto \[8px] &= 2\dfrac\left( x^3\right) – 4\dfrac\left( x^2\right) \dfrac(x) \[8px] &= 2 \left(3 x^2 \right) – 4 \left(2x^1 \right) 1 \[8px] &= 6x^2 -8x 1 \quad \cmark \end$ Recall that $\dfrac\left(x^n\right) = nx^.$ The rule also holds for fractional powers: $\begin \dfrac\left(\sqrt \right) &= \dfrac\left(x^ \right) \[8px] &= \fracx^ \[8px] &= \fracx^ \quad \cmark \[8px] &= \frac\frac \quad \cmark \end$ Note that the last two lines are completely equivalent, and either would be acceptable as the answer.Recall that $\dfrac\left(x^n\right) = nx^.$ The rule also holds for negative powers: $\begin \dfrac \left(\dfrac \right) &= 5 \dfrac \left(x^ \right) \[8px] &= 5 \left((-3) x^ \right) \[8px] &= -15 x^ \quad \cmark \[8px] &= \frac \quad \cmark \end$ Note that the last two lines are completely equivalent, and either would be acceptable as the answer.first have to develop the function you’re going to maximize or minimize, as we did in Stage I above.Having done that, the remaining steps are exactly the same as they are for the max/min problems you recently learned how to solve.Notice, by the way, that so far in our solution we haven’t used any Calculus at all.That will always be the case when you solve an Optimization problem: you don’t use Calculus until you come to Stage II.We could reason physically, or use the First Derivative Test, but we think it’s easiest in this case to use the Second Derivative Test.Let’s quickly compute the second derivative, starting with the first derivative that we found above:$\begin A'(r) &= 4\pi r\, – 2V r^ \[8px] A’^\prime(r) &= 4\pi \dfrac(r) -2V \dfrac\left(r^ \right) \[8px] &= 4\pi -2V \left((-2)r^ \right) \[8px] &= 4\pi \frac \end$Since $r (as opposed to a maximum or saddlepoint), which is what we’re after: The minimum surface area occurs when$r = \sqrt[3]\,. Now that we’ve found the critical point that corresponds to the can’s minimum surface area (thereby minimizing the cost), let’s finish answering the question: The problem asked us to find the dimensions — the radius and height — of the least-expensive can.

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