Download Problems in Finding Derivatives and Tangent Lines SOLUTION and more Study notes Calculus in PDF only on Docsity!
Problems in Finding Derivatives and Tangent Lines
SOLUTION
- Use the definition of the derivative to compute f ′(1) for f (x) =
x + 7
. NO
CREDIT will be given for any other method.
Solution:
f ′(1) = limh→ 0
f (1 + h) − f (1) h
= limh→ 0
1 + h + 7 −^
h = limh→ 0
13 · 8 − 13(8 + h) 8 h(8 + h) = limh→ 0
− 13 h 8 h(8 + h) = limh→ 0
8(8 + h)
- Compute the derivatives of the following functions. DO NOT SIMPLIFY.
(a) f (x) = (sinx + x^2 + 1)(2x^3 + x)^2
(b) g(x) =
x^2 + 12x + e^3 x + ex
(c) h(t) = e^3 t(t^2 + x^2 )
Solution:
(a) f ′(x) = (cosx + 2x)(2x^3 + x)^2 + 2(sinx + x^2 + 1)(2x^3 + x)(6x^2 + 1)
(b) g′(x) =
(2x + 12)(x + ex) − (x^2 + 12x + e^3 )(1 + ex) (x + ex)^2
(c) h′(t) = 3e^3 t(t^2 + x^2 ) + e^3 t(2t)
- Find h′(1) where h(x) =
xg(x) + 7 f (x)
, f ′(1) = 4, g′(1) = −2, and f (1) = 1,
g(1) = 1. EXPREE YOUR ANSWER AS AN INTEGER.
Solution:
h′(x) =
f (x)(xg(x) + 7)′^ − (xg(x) + 7)f ′(x) [f (x)]^2
=
f (x)[g(x) + xg′(x)] − [xg(x) + 7]f ′(x) [f (x)]^2
∴ h′(1) =
f (1)[g(1) + g′(1)] − [g(1) + 7]f ′(1) [f (1)]^2 =
1[1 + (−2)] − [1 + 7]
[1]^2
- Find the equation of the tangent line to y = f (x) =
x + 3 2 x + 1
at the point
corresponding to x = 0.
Solution:
The slope of the tangent line at x = 0 is f ′(0).
f ′(x) =
(2x + 1) − 2(x + 3) (2x + 1)^2
∴ f ′(0) =^1 −^6 1
Therefore, the equation of the tangent line is: y = − 5 x + b.
At x = 0, y = f (0) =
= 3. Therefore 3 = 0 + b, i.e. b = 3.
The equation of the tangent line is: y = − 5 x + 3.
