2.2 Derivative of Polynomial Functions - La Citadelle

1 Calculus and Vectors How to get an A+ Derivative of Polynomial Functions 2010 Iulia & Teodoru Gugoiu - Page 1 of 4 Derivative of Polynomial Functions A Power Rule Consider the power function: Rnxxyn =,,. Then: 1)'( =nnnxx 1 =nnnxxdxd Some useful specific cases: xxx21)'(1)'(0)'1(=== Ex 1. For each case, differentiate. a) 00)'()'1(100=== xx b) 1)(1)(1)'()'(0111==== xxxx c) 41555)(5)'(xxx== d) 22111'1)1()'(1xxxxx = = == e) 88177'777)7()'(1xxxxx = = == f) xxxxxx211212121)'()'(212112121===== g) 32323213131331313131)'()'(xxxxxx===== h) 44141143434343434343)'()'(xxxxxx===== i) 1)'( = xx B Constant Function Rule Let consider a constant function: Rccxf =,)(. Then: 0)'(=c 0=cdxd Ex 2. Find each Derivative function: a) 0)'2(= b) 0)'(= C Constant Multiple Rule Let consider )()(xcfxg=. Then: ')'()()]([)(')]'([cfcfxfdxdcxcfdxdxcfxcf === Ex 3. Differentiate each expression: a) 32x 213336)3)(2()')(2()'2(xxxx = = = b) 23x 331222'266)2)(3()')(3()'3(3xxxxxx== = = = c) x2 xxxx21212)'(2)'2(=== D Sum and Difference Rules '')'()()()]()([)(')(')]()([gfgfxgdxdxfdx dxgxfdxdxgxfxgxf = = = Ex 4.

2 Differentiate. a) 723432)(xxxxf + = 6667272722183)('2183)7(3)2(43)'(3)'(4)') (3(0)'3()'4()'3()'2()'3432()('xxxfxxxxxx xxxxxxxxf + = + = + = + += ++ += + = Calculus and Vectors How to get an A+ Derivative of Polynomial Functions 2010 Iulia & Teodoru Gugoiu - Page 2 of 4 b) 52421)(xxxxg+ = 632632632521521'522041)('2041)5(4)2)(2() (1)'4()'2()'()'42(421)('xxxxgxxxxxxxxxxx xxxxxg + = + = + + =+ +=+ = + = c) xxxxh3)( = 3232353352313212132213221131121'3'33221) ('3221322132213221)'()'()'()'()('xxxxxhx xxxxxxxxxxxxxxxxxxxxxxxh+ = + =+ = = = = = = = = E Tangent Line To find the equation of the tangent line at the point ))(.(afaP: 1. Find Derivative function )('xf. 2. Find the slope of the tangent line using: )('afm= 3. Use the slope-point formula to get the equation of the tangent line: )()(axmafy = Ex 5. Find the equation of the tangent line at the point )0,1(Pto the graph of xxxfy1)(2 ==.

3 33)1(3031/1)1(2)1(12)('22 = = =+==+=xyxyfmxxxf Ex 6. Find the equation of the tangent line of the slope 2=mto the graph of xxfy2)(==. Graph the function and the tangent line. 2/12)4/1(21)1,4/1(141241212)('1212)('+= = == = = ===xyxyPyxxxfxxxf 2 112 2 112xyy = 2*sqrt(x)y = 2*x+1/2 Ex 7. Find the points on the graph of 132)(23+ ==xxxfy where the tangent line is horizontal. mxfxxxxxf= = =)(')1(666)('2 0=m (horizontal tangent) )0,1(01)1,0(100)1(6 ByxAyxxx = = = == The tangent line is horizontal at )1,0(Aand )0,1(B. 3 2 1123 3 2 1123xy Calculus and Vectors How to get an A+ Derivative of Polynomial Functions 2010 Iulia & Teodoru Gugoiu - Page 3 of 4 F Normal Line If Tm is the slope of the tangent line, then slope of the normal line Nm is given by: TNmm1 = Ex 8. Find the equation of the normal line to the curve xxxfy2)(+== at )3,1(P. 2)1(13111)1(21)('2+= = = = == =xyxymmfmxxfTNT Ex 9. Analyze the differentiability of each function. a) 3)(xxfy== The function f is continuous over R.

4 323213131313131)'()('xxxxxf==== )('xfdoes not exist at 0=x. Therefore the function fis not differentiable at 0=x. As 0 x, )('xf. The point )0,0(Ois a infinite slope point. 3 2 1123 3 2 1123xy b) 3/2)(xxfy== The function f is continuous over R. 331132323232)('xxxxf=== )('xfdoes not exist at 0=x. The function fis not differentiable at 0=x. As 0x, )('xf. As + 0x, )('xf. The point )0,0(Ois a cusp point. 3 2 1123 3 2 1123xy c) |3|)( ==xxfy < =3,33,3)(xxxxxf < >= 3,13,1)('xxxf 1)('lim1)('lim33= =+ xfxfxx 'f does not exist at 3=x. Therefore, fis not differentiable at 3=x. The point)0,3(Pis a corner point (see the figure to the right side). 2 11234567 4 3 2 11234xy Calculus and Vectors How to get an A+ Derivative of Polynomial Functions 2010 Iulia & Teodoru Gugoiu - Page 4 of 4 H Differentiability for piece-wise defined Functions Let consider the piece-wise defined function: >=<=axxfaxcaxxfxf),(,),()(21 The function f is differentiable at ax=if: (a) the function is continuous at ax= (b) )(')('21afaf=(the slope of the tangent line for the left branch is equal to the slope of the tangent line for the right branch).

5 Ex 10. Analyze the differentiability of each function at 1=x. a) > =1,21,)(2xxxxxf 1)1(,2)(lim,1)(lim11===+ fxfxfxx The function f is not continuous at 1=x. Therefore, the function f is not differentiable at 1=x. See the figure below. 3 2 1123 11234xy b) a) > =1,1,)(2xxxxxf 1)1(,1)(lim,1)(lim11===+ fxfxfxx The function is continuous at 1=x. > <=1,11,2)('xxxxf 1)(lim,2)('lim11==+ xfxfxx The function f is not differentiable at 1=x (corner point). See the figure below. 3 2 1123 11234xy c) a) > =1,121,)(2xxxxxf 1)1(,1)(lim,1)(lim11===+ fxfxfxx The function is continuous at 1=x. > <=1,21,2)('xxxxf 2)(lim,2)('lim11==+ xfxfxx The function f is differentiable at 1=x. See the figure below. 3 2 1123 2 112345xy Reading: Nelson Textbook, Pages 76-81 Homework: Nelson Textbook: Page 82 #2cdf, 3ce, 4aef, 5b, 6b, 7a, 8b, 9b, 11, 14, 17, 20, 25iii, 28