ANALYTIC GEOMETRY AND CALCULUS

Running head: ANALYTIC GEOMETRY AND CALCULUS I
ANALYTIC GEOMETRY AND CALCULUS I
Name of the Student
Name of the University
Author Note

ANALYTIC GEOMETRY AND CALCULUS I1
Question 1:
a) Derivative of a function f(x) is defined by the rate of change of f(x) with respect to x. If the
derivative at a point is negative then the function is decreasing and if the derivative at a point
if positive then the function is increasing at that point. By limit theorem the derivative of f(x)
with respect x is given by,
lim
h→ 0
f ( x +h ) −f ( x )
h
b) d
dx (4 x2 +5 x ) = lim
h→ 0
4 ( x+ h ) 2+ 5 ( x +h ) −(4 x2 +5 x )
h
= lim
h→ 0
4 x2+ 8 xh+ h2+ 5 x +5 h−( 4 x2+5 x )
h
= lim
h→ 0
8 xh+h2+5 h
h
= lim
h→ 0
8 x +h+ 5 = 8x+ 5 (Proved)
c) The equation of line tangent to f(x) = 4x^2 + 5x at a point (1,9) is given by,
(y – 9)/(x-1) = m (where m = slope of the line)
Now, m = slope of the line = gradient at (1,9) = 8*1 + 5 = 13.
Hence, the equation of the line is
(y – 9)/(x-1) = 13
 y – 9 = 13x – 13
 y = 13x + 4

ANALYTIC GEOMETRY AND CALCULUS I2
Question 2:
1.
d
dx ( x5 sin (5 x)) = 5 x4∗sin (5 x ) +5 x5∗cos ( 5 x )
(Using product rule of differentiation and d/dx(sin(mx)) = mcos(mx))
2.
d
dx ( cosx
x2+ 4 x ) = − ( x2+ 4 x ) sinx−cosx (2 x +4 )
( x2 + 4 x )2 = −x2 sinx−4 xsinx−2 x cosx −4 cosx
( x2 + 4 x )
2 (using
quotient rule of differentiation d ( u
v )= vdu – udv
v2
3.
d
dx (5 x7 – x−8 +arctan (7 x)) = 35 x6+ 8 x−9+ 7
( 7 x ) 2+ 1 = 35 x6+ 8 x−9+ 7
49 x2 +1
(By using formula d/dx(x^n) = nx^(n-1) and d(arctan (7 x)¿ = 7/(1+49x^2))
4.
d
dx ( √ sinx+23 x +5)
= ( ½ )∗( sinx+23 x+5 )
−1
2 ( cosx+ 23 ) (By chain rule and
Question 3:
a)
Given equation of the line is
y=4 e5 x cosx