MATH 1A - MIDTERM 2 - SOLUTIONS. f(x) = x. f f(x) f(4) (4) = lim. x 4. x 2 = lim. x 4. = lim. x 4 x + 2. = lim. 1 = lim = = 1 4

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1 MATH 1A - MIDTERM 2 - SOLUTIONS PEYAM RYAN TABRIZIAN 1. (15 points) Using the definition of the derivative, calculate f (4), where: f() = f f() f(4) (4) ( 2)( + 2) 4 ( 4)( + 2) 4 4 ( 4)( + 2) = = 1 4 Hence, f (4) = 1 4 Date: Friday, July 15th,

2 2 PEYAM RYAN TABRIZIAN 2. (15 points) Using the definition of the derivative, calculate f (), where: f() = 2 + f f() f(a) (a) a a 2 + (a 2 + a) a a a 2 + a 2 a a 2 a 2 + ( a) a a 2 a 2 a a + a a ( a)( + a) a a + a + lim 1 a a =2a lim a a a Hence, f () = 2 + 1

3 Other solution: MATH 1A - MIDTERM 2 - SOLUTIONS 3 f (a) f(a + h) f(a) h (a + h) 2 + (a + h) (a 2 + a) h a 2 + 2ah + h 2 + a + h a 2 a h 2ah + h 2 + h h h(2a + h + 1) h 2a + h + 1 =2a + 1 Hence, f () = 2 + 1

4 4 PEYAM RYAN TABRIZIAN 3. (50 points, 5 points each) Find the derivatives of the following functions: (a) f() = e + cos() + 1 f () = e sin() (b) f() = ln() f () = ln() + ( ) 1 1 = ln() = ln() (c) f() = e (sin()) 2 f () = e (sin()) 2 e 2 sin() cos() (sin()) 4 = e (sin() 2 cos()) (sin()) 3 (d) f() = ln( 2 + 1) ( f () = 2 1 ln( 2 +1) ) ( ) (2) Note: Woe to you if you tried to simplify! ln( 2 + 1) ln( 2 ) + ln(1) (e) f() = tan(tan(tan())) f () = sec 2 (tan(tan())) sec 2 (tan()) sec 2 ()

5 (f) f() = (sin()) MATH 1A - MIDTERM 2 - SOLUTIONS 5 Let y = (sin()) 1) ln(y) = ln(sin()) 2) y cos() = ln(sin()) + y sin() ( ) 3) y = y ln(sin()) + cos() sin() y = (sin()) ( ln(sin()) + cos() sin() ) (g) y, where 2 + 3y + y 2 = 1 Differentiating: 2 + 3y + 3y + 2yy = 0 Solving for y : y (3 + 2y) = (2 + 3y) y = (2 + 3y) 3 + 2y (h) The equation of the tangent line to y = 4 +3 at the point (1,4) y = Slope = y (1) = = 7 Equation: y 4 = 7( 1) (i) The equation of the tangent line at (1, 0) to the curve: sin(y 2 + sin(y)) = Differentiating: cos(y 2 + sin(y))(2yy + sin(y) + cos(y)y ) = 2

6 6 PEYAM RYAN TABRIZIAN If = 1 and y = 0, we get: cos(0)( y ) = 2, so y = 2 Equation: y 0 = 2( 1), so y = 2 2 (j) f (), where f() = tan 1 () f () = f () = 1 (2) = (1+ 2 ) 2 2 (1+ 2 ) 2

7 MATH 1A - MIDTERM 2 - SOLUTIONS 7 4. (20 points) Show that the sum of the and y intercepts of any tangent line to the curve + y = c is c. Slope: ( ) y 2 =0 y ( ) 1 y 2 = 1 y 2 y = y y = 2 y 2 y y = Equation: At ( 0, y 0 ), the slope is y 0 0, and so the equation of the tangent line at ( 0, y 0 ) is: y0 y y 0 = ( 0 ) 0 y intercept: To find the y intercept, set = 0 and solve for y: y0 y y 0 = (0 0 ) 0 y0 y y 0 = ( 0 ) 0 y y 0 = y 0 0 y =y 0 + y 0 0 intercept: To find the intercept, set y = 0 and solve for :

8 8 PEYAM RYAN TABRIZIAN y0 0 y 0 = ( 0 ) 0 y0 y 0 = ( 0 ) = ( y 0 ) y0 = y0 Sum: The sum of the y and intercepts is: (y 0 + y 0 0 ) + ( y0 ) = y0 + y 0 Using the hint: y0 + y 0 = ( 0 ) y0 + ( y 0 ) 2 = ( 0 + y 0 ) 2 But since ( 0, y 0 ) is on the curve + y = c, we get 0 + y0 = c. And so, finally we get that the sum of the and y intercepts is: y0 + y 0 = ( 0 + y 0 ) 2 = ( c) 2 = c

9 MATH 1A - MIDTERM 2 - SOLUTIONS 9 Bonus 1 (5 points) Assume f is a nonzero function which satisfies: { f () = f() f(0) = 1 For this problem, the following property might be useful: Property: If g () = 0 for all, then g() = C, where C is a constant. (a) Show that f( + a) = f()f(a). Hint: Define g() = f(+a) f() g () = f ( + a)f() f( + a)f () (f()) 2 = f( + a)f() f( + a)f() (f()) 2 = 0 (We used the fact that f ( + a) = f( + a) and f () = f()) So g () = 0, so g() = C. To find out what C is, notice that: C = g(0) = f(a) f(0) = f(a) Hence C = f(a), and we get that g() = f(a), so f( + a) = f(a) f() And multiplying by f(), we get: f( + a) = f()f(a)

10 10 PEYAM RYAN TABRIZIAN (b) Show that f( ) = 1 f() Hint: Define g() = f( )f(). g () = f ( )f() + f( )f () = f( )f() + f( )f() = 0 Hence g() = C. But: C = g(0) = f(0)f(0) = 1 1 = 1 So g() = 1, and f( )f() = 1, and: f( ) = 1 f()

11 MATH 1A - MIDTERM 2 - SOLUTIONS 11 (c) Show that f(a) = f() a Hint: Define g() = f(a) f() a g () = f (a)a(f()) a f(a)af() a 1 f () f() 2a = af(a)f()a af(a)f() a 1 f() f() 2a = af(a)f()a af(a)f() a f() 2a =0 So g() = C, but: So g() = 1, so f(a) f() a C = g(0) = f(0) f(0) a = 1 1 = 1 = 1, so: f(a) = f() a

12 12 PEYAM RYAN TABRIZIAN (d) (2 etra points) In fact, the converse statement is true too! Namely, if f is a function with: Show that f() = e. f(a + b) = f(0) = 1 f (0) = 1 f(a)f(b) Notice that we re not making any assumptions about f being smooth, only that it is differentiable at 0! Hint: All you need to show is that f () = f() for all. f () f( + h) f() h f()f(h) f() h f() f(h) 1 h f(h) 1 =f() lim =f()f (0) =f() h

13 MATH 1A - MIDTERM 2 - SOLUTIONS 13 Bonus 2 (5 points) The following bonus problem is meant to show you that derivatives can behave in very strange ways! (a) Find an eample of a function f with lim f() = 0 but lim f () 0 (in fact, the limit does not eist) Let f() = sin(2 ). Then lim f() = 0 by the Squeeze Theorem. But: f () = cos(2 )(2) sin( 2 ) = 2 cos( 2 ) sin(2 ) 2 2 Now lim sin( 2 ) 2 = 0 by the squeeze theorem, but lim cos( 2 ) does not eist, from which it follows that: lim f () does not eist! (so in fact it is 0)

14 14 PEYAM RYAN TABRIZIAN (b) Find an eample of a function f that is differentiable at 0, but whose derivative is not continuous at 0. Let f() = 2 sin( 1 ). Then: f (0) 0 f() f(0) 0 f () = 2 sin (by the squeeze theorem) 0 2 sin( 1 ) 0 So f is differentiable at 0 with f (0) = 0. However: ( ) 1 + ( 2 1 ) cos 2 ( ) 1 = 2 sin ( ) 1 sin = 0 0 ( ) 1 cos ( ) 1 Now lim 0 2 sin( 1) = 0 (by Squeeze Thm), but lim 0 cos ( ) 1 does not eist, so lim 0 f () does not eist, and in fact does not equal to f (0) = 0. So lim 0 f () f (0), hence f is not continuous at 0.