The Chain Rule solved questions

CHAPTER 9
The Chain Rule

9.1 If f(x) = x2 + 2x — 5 and g(x) = x3, find formulas for the composite functions /°g and g°/.
(/°g)W=/(g(*))=/(* 3 ) = (*3)2 + 2(* 3 )-5 = * 6 +2* 3 -5
(g°/)« = S(/M) = 8(*2 + 2x-5) = (x2 + 2x- 5)3

9.2 Write the function is the composition of two functions.
Let g(x) = 3x - 5 and let f(x) = Vx. Then ( f°*)<*) = /(gW) = A3* - 5) = V3l^5.

9.3 If f(x) = 2x and g(x) = \/(x- 1), find all solutions of the equation (f°g)(x) = (g°/)(x).

(/•*)(*)= A *(*))=/ and (*•/)(*) = *( A*)) = *(2*) =

So, we must solve 4x-2 = x-l, 3x = l, * = 5 . Answer

9.4 Write the chain rule formula for the derivative of f» g

(/"*)'(*) =/'(*<*))•*'(*)•

9.5 If y = F(u) and M = G(x), then we can write y = F(G(x)). Write the chain rule formula for dyldx,
where we think of y as a function of x.

Here, the first occurrence of y refers to y as a function of AC, while the second occurrence of y
(on the right side) refers to y as a function of «.

9.6 Find the derivative of (*3 - 2x2 +7x- 3)4.
Use the chain rule. Think of the function as a composition (f°g)(x), where f(x) = x4 and g(x) =
x3 - 2x2 + 7x-3. Then f '(x) = 4x3 and g'(jr) = 3* -4x + 7. Hence, x3 - 2x2 + 7x- 3)4 = 4(x3 -
2 3 2
2x + 7x- 3) • (3x -4x + 7).

9.7 Find the derivative of

We can write = (3*2 + 5)-4. Now we can use the chain rule. Remember that x'^rr'- 1
for any real number r. In particular, »-« = -4x-5. By the chain rule,

(3x2 + 5)"4 = -4(3x2 + 5K5 • (6jr) = -


9.8 Find the derivative of
We can write = (2x + 7)1'2. By the chain rule,



Here, we have used and


9.9 Find the derivative of

56

, THE CHAIN RULE 57


Use the chain rule. Here, we must calculate by
the quotient rule: Hence,

Answer

9.10 Find the derivative of (4x2 - 3)2(x + 5)3.
Think of this function as a product of (4x2 - 3)2 and (x + 5)3, and first apply the product rule:
[(4x2-3)2(jt + 5)3] = (4;c 2 -3) 2 - (x + 5)3 + (x + 5)3 • (4*2-3)2. By the chain rule, (* + 5)3 =
2 2 2 2 2 2
3(* + 5) • 1 = 3(* + 5) , jnd (4x - 3) = 2(4* - 3) • (8x) = I6x(4x - 3). Thus, t(4^ 2 -3) 2 (jc +
3 2 2 2 3 2 2
5) ] = (4^ - 3) • 3(x + 5) + (x + 5) -16^(4jc -3). We can factor out (4x - 3) and (x + 5)2 to ob-
tain: (4^;2 - 3)(x + 5)2[3(4*2 - 3) +16^ + 5)] = (4x2 - 3)(x + 5f(12x2 - 9 + 16x2 + 80x) = (4x 2 -3)(* +
5)2(28x2 + 80A:-9). Answer

9.11 Find the derivative of

The chain rule is unnecessary here. So,

Also, So, Thus,



9.12 Find the derivative of

By the chain rule, But,
Hence,




9.13 Find the slope-intercept equation of the tangent line to the graph of at the point (2, 5)-

By the quotient rule,

By the chain rule,


Thus,
When x = 2, and, therefore, at the point (2, 5), Hence, a point-slope equation of
the tangent line is Solving tor y, we obtain the slope-intercept equation

9.14 Find the slope-intercept equation of the normal line to the curve it the point (3,5).
2 2
3, = (* + 16)" . Hence, by the chain rule,



At the point (3,5), and, therefore. This is the slope of the tangent line. Hence, the
slope of the normal line is — § , and a point-slope equation for it is Solving for y, we obtair
the slope-intercept equation

9.15 If y = x —2 and x = 3z2 + 5, then y can be considered a function of z. Express in terms of z.