MATH235 Calculus 1
Problems on Finding Derivatives.
09/23/2010
Find the derivative of the given function.
1. y = 9 tan
x
3
dy
x 1
= 9 sec2 ( )
dx
3 3
2 x
= 3 sec ( )
3
2 2 θ
2 θ 2
2. s = sin (θ e ) = (sin(θ e ))
by the Chain Rule,
ds
d
= 2 sin(θ2 eθ ) (sin(θ2 eθ ))
dθ
dθ
d
= 2 sin(θ2 eθ ) cos(θ2 eθ ) (θ2 eθ )
dθ
2 θ
2 θ
= 2 sin(θ e ) cos(θ e )(2θeθ + θ2 eθ )
by the Power Rule and the Chain Rule,
by the Chain Rule,
by the Product Rule,
= 2θeθ sin(θ2 eθ ) cos(θ2 eθ )(2 + θ)
3. y = (9x2 − 6x + 2)ex
3
dy
d 3
3
= (18x − 6)ex + (9x2 − 6x + 2) ex
dx
dx
3
x3
2
= (18x − 6)e + (9x − 6x + 2)ex 3x2
x3
4
3
by the Product Rule,
by the Chain Rule,
2
= e (27x − 18x + 6x + 18x − 6)
√
1
4. r = sec θ tan
θ
√
√ d
dr
d
1
1
= (sec θ) tan + sec θ (tan )
dθ
dθ
θ
dθ
θ
√
√ d√
√
1
1 d 1
= sec θ tan θ(
θ) tan + sec θ sec2 (
)
dθ
θ
θ dθ θ
√
√
√
1
1
1 −1
= sec θ tan θ( √ ) tan + sec θ sec2 ( 2 )
θ
θ θ
2 θ
√
√
1
1
1
1
= sec θ( √ tan θ tan − 2 sec2 )
θ θ
θ
2 θ
1
by the Product Rule,
by the Chain Rule,
2
5. y = e
√
3x+1
√
dy
d√
= e 3x+1
3x + 1
dx
dx
√
1 d
1
= e 3x+1 (3x + 1)− 2 (3x + 1)
2
dx
√
3x+1
3e
= √
2 3x + 1
6. y = sin5 x = (sin x)5
by the Chain Rule,
by the Power Rule and the Chain Rule,
dy
d
= 5 sin4 x sin x
dx
dx
= 5 sin4 x cos x
7. y =
by the Chain Rule,
2
= 2(3x − 2)−1
3x − 2
dy
d
= −2(3x − 2)−2 (3x − 2)
dx
dx
−6
=
(3x − 2)2
by the Power Rule and the Chain Rule,
2
8. y = 3x(x2 + 2x) 3
2
2
dy
d
= 3(x2 + 2x) 3 + 3x ((x2 + 2x) 3 )
dx
dx
2
1 d
2
= 3(x2 + 2x) 3 + 3x( (x2 + 2x)− 3 (x2 + 2x))
3
dx
2
2
4x
+
4x
= 3(x2 + 2x) 3 +
1
(x2 + 2x) 3
3(x2 + 2x) + 4x2 + 4x
=
1
(x2 + 2x) 3
7x2 + 10x
=
1
(x2 + 2x) 3
by the Product Rule,
by the Power Rule and the Chain Rule,