Calculus 220 BD1
Quiz 5 (10 points) Solutions
SL030112
1. Find the derivatives of the following. You need not simplify algebraically. Use proper
notation.
1
(a) (2 points) f (x) = p
cot(x2 ) + 10
First, rewrite the function as
f (x) = (cot(x2 ) + 10)−1/2
The outside function is
−1/2
f 0 (x) =
, so we use the power rule and chain rule:
−3/2
−1
cot(x2 ) + 10
· (cot(x2 ) + 10)0
2
To compute the remaining derivative, we have to use the chain rule again, where the
outside function is now cot, so we get
f 0 (x) =
−3/2
−1
cot(x2 ) + 10
· − csc2 (x2 ) · (x2 )0
2
The remaining derivative just requires the power rule:
f 0 (x) =
−3/2
−1
cot(x2 ) + 10
· − csc2 (x2 ) · 2x
2
No need to simplify.
(b) (2 points) g(θ) = 7sin−1 (2θ − 5) + 3csc(θ)
We need the chain rule for the derivative:
1
g 0 (θ) = 7 · p
· (2θ − 5)0 + 3csc(θ) ln(3) · (csc(θ))0
2
1 − (2θ − 5)
so computing the remaining derivatives gives
1
· 2 + 3csc(θ) ln(3) · − csc(θ) cot(θ)
g 0 (θ) = 7 · p
2
1 − (2θ − 5)
Again, no need to simplify.
2. (3 points) Find d2 y/dx2 if y = tan(4x).
First, compute
dy
= sec2 (4x) · 4
dx
using the chain rule. Now compute
d2 y
d dy
d
d
=
=
4 sec2 (4x) =
4 (sec(4x))2
2
dx
dx dx
dx
dx
We need the chain rule:
d2 y
= 8(sec(4x))·(sec(4x))0 = 8 sec(4x)·sec(4x) tan(4x)·(4x)0 = 8 sec(4x)·sec(4x) tan(4x)·4
dx2
1
Calculus 220 BD1
Quiz 5 (10 points) Solutions
SL030112
3. (3 points) Find the equation of the line tangent to the curve
sin(y) + 4 cos(x) = 8y
at the point (π/2, 0).
We want a tangent line. The slope of tangent lines is given by dy/dx, so let’s use implicit
differentiation:
dy
dy
cos(y) − 4 sin(x) = 8
dx
dx
We could solve for dy/dx right away, but in this case (since we have to plug in a point
eventually) let’s plug in right away. We get:
cos(0)
dy
dy
dy
dy
−4
dy
dy
− 4 sin(π/2) = 8
=⇒
−4=8
=⇒ −4 = 7
=⇒
=
dx
dx
dx
dx
dx
7
dx
and thus our line is
y−0=
−4
(x − π/2)
7
2