calculus question

Question

find the derivatives of the given functions. show all work and simplify the answer if possible.

f(x) = (sinx-xcosx)/cosx+xsinx

Answer 1

f(x) = (sinx-xcosx)/cosx+xsinx

 

*** (assuming your brackets are correct, and the "/cosx + xsinx" is not bracketed, meaning xsinx is by itself)

we can rewrite the divisor, cosx cosx^-1 in the numerator

f = (sinx - xcosx)*(cosx^-1) + xsinx

apply product rule     h'(x)=[f(x)g(x)]'= f'(x)g(x)+f(x)g'(x)    where f'(x) = df(x)/dx

---------------------

 

[remember:  d/dx (sinx) = cosx      and      d/dx (cosx) = -sinx ]

d/dx [ (sinx - xcosx)*(cosx^-1) ] + d/dx [xsinx]

d/dx(sinx - xcosx)*(cosx^-1) + (sinx - xcosx)*d/dx(cosx^-1) + d/dx(x)*sinx + x*d/dx(sinx)
.............#1........................   ................ #2 ....................   .................. #3.................

#1
[ cosx - (d/dx(x)*cosx) + x*d/dx(cosx)) ] * cos^-1      (applied prod rule on xcosx)
[ cosx - [1*cosx + x*(-sinx) ] * cosx^-1
[cosx - cosx + xsinx ] * cosx^-1

= xsinx / cosx = xtanx

#2 (sinx - xcosx)*(-1)(cosx^-2)(-sinx) = (sinx - xcosx)*sinx*cosx^-2
(sinx^2 - xcosx*sinx) * (cosx^-2)
(sinx^2/cosx^2) - (xcosx*sinx)/cosx^2

= tanx^2 - x*sinx/cosx = tanx^2 - xtanx

#3 (1)*sinx + x*cosx
= sinx + xcosx

now combine #1,2,3
-------------------------

xtanx + (tanx^2 - xtanx) + (sinx + xcosx)    ..................  (xtanx terms cancel)

= tanx^2 + sinx + xcosx

Answer 2

Hi, Ashley!

 

To find the derivative of this function, there are a couple of derivative rules you need to know:

1. Product Rule:  Part1 * (derivative of Part2) + (derivative of Part1) * Part2

2. Quotient Rule:  ho d hi - hi d ho

                               hoho

That is....  denominator * (derivative of numerator) - numerator * (derivative of denominator)

denominator²

3. Derivatives of trig functions:

d(sinx) = cosx

d(cosx) = -sinx

d(-sinx) = -cosx

d(-cosx) = sinx

 

Now, f(x) = sinx - xcosx

                 cosx + xsinx

Step 1: Start by asking yourself "What math function holds this whole function together?"  In this case, the division line connects all parts of this function, so this is the part we need to begin with.  To find the derivative of a quotient, we need to use the quotient rule: "ho d hi minus hi d ho all over hoho"  (it sounds crazy, but it's a great memorization tool!  "ho" sounds like low and means denominator.  "hi" is like high and stands for the numerator.  "d" simply means "derivative of".).

"ho                     d      hi           -        hi              d         ho              all over              

(cosx + xsinx) * d(sinx - xcosx) - (sinx - xcosx) * d(cosx + xsinx)         ↵     hoho"

                  (cosx + xsinx) * (cosx + xsinx)                                                  ↵

 

Step 2: In order to find the derivative of the numerator (or denominator), we will need to apply 2 derivative concepts, one of which is a rule that I listed above.  First off, we ask ourselves the same question as in step 1: "What math function holds the whole numerator together?"  Answer: Subtraction.  Finding the derivative of something with subtraction is simple.  You just take the derivative of each part as if they were separate problems.  But look at the second part of the subtraction.  It contains multiplication, and so we need to apply the product rule when we take the derivative of the second part.  Let's use the derivative of the numerator as an example:

d(sinx - xcosx) = d(sinx) - d(xcosx)

= cosx - (x*-sinx + cosx)

= cosx - -xsinx - cosx

= xsinx

If you apply the same rules and process to the derivative of the denominator, you will get:

d(cosx + xsinx) = -sinx + xcosx + sinx

= xcosx

 

Step 3:  Plug everything in and simplify.

(cosx + xsinx) * d(sinx - xcosx) - (sinx - xcosx) * d(cosx + xsinx)
(cosx + xsinx) * (cosx + xsinx)

 

(cosx + xsinx)*xsinx - (sinx - xcosx)*xcosx

(cosx + xsinx)²

 

= xcosxsinx + x²sin²x - (xsinxcosx - x²cos²x)

        (cosx + xsinx)²

= xcosxsinx + x²sin²x - xsinxcosx + x²cos²x

       (cosx + xsinx)²

= x²sin²x + x²cos²x

       (cosx + xsinx)²

= x² (sin²x + cos²x)      ←Note: sin²x + cos²x = 1

      (cosx + xsinx)²

= x²                  

  (cosx + xsinx)²

 

I think this is as simplified as it can get.  I hope this was helpful!

 

~ Patty S.