Calculus: The Language of Change

Exercise Set 6.7
Combined Symbolic Rules

by Sharon Portz

Exercise 1 : All rules apply here

Use all the rules of differentiation to find the following:

a)  y = (1/(2 x  -  1))			b) y = (1/(a x  +  b)) ( u = ax + b; y = (1/u) = (u^(-1)))

c) y = Cos[(x^2)]			d) y = (Cos^2)[x]

e)  y = (Sqrt[1 - (x^2)])		f) y = (1/(Sqrt[1 - (x^2)]))

g) y = (((12 - 3 (x^7)))^8)		h) y = Sin[(x^2) + (x^3)]

i) y = (2 (x^3) + 4)((x^2) - (Sqrt[x]))	j) y = (([(((a x  +  b))^(-1)) + c])^(-1))

k) y = Sin[x]Cos[x]		l) y = Sin[2x]

m) y = Sin[3x]			n) y = (1/Cos)[3  x]

o) y = Sin[x] (([Cos[x]])^(-1)) = Sin[x]/Cos[x] = Tan[x]
p) y = ((2  -  3  x)/(1 +  2  x)) = (2 - 3x)[(((1  +  2  x))^(-1))]

part a

a)  y = (1/(2 x  -  1))

Solution of part a

part b

b) y = (1/(a x  +  b)) ( u = ax + b; y = (1/u) = (u^(-1)))

Solution of part b

part c

c) y = Cos[(x^2)]

Solution of part c

part d

d) y = (Cos^2)[x] (unchain two ways)

Solution of part d

part e

e)  y = (Sqrt[1 - (x^2)])

Solution of part e

part f

f) y = (1/(Sqrt[1 - (x^2)]))

Solution of part f

part g

g) y = (((12 - 3 (x^7)))^8)

Solution of part g

part h

h) y = Sin[(x^2) + (x^3)]

Solution of part h

part i

i) y = (2 (x^3) + 4)((x^2) - (Sqrt[x]))

Solution of part i

part j

j) y = (([(((a x  +  b))^(-1)) + c])^(-1))

Solution of part j

part k

k) y = Sin[x]Cos[x]

Solution of part k

part l

l) y = Sin[2x]

Solution of part l

part m

m) y = Sin[3x]

Solution of part m

part n

n) y = (1/Cos)[3  x]

Solution of part n

part o

o) y = Sin[x] (([Cos[x]])^(-1)) = Sin[x]/Cos[x] = Tan[x]

Solution of part o

part p

p) y = ((2  -  3  x)/(1 +  2  x)) = (2 - 3x)[(((1  +  2  x))^(-1))]

Solution of part p

Exercise 2 : (Sqrt[(x^2) + 2 x + 1])

a) Differentiate y = (Sqrt[(x^2)+ 2 x + 1]) using the Chain Rule, the Power Rule, and the Superposition Rule.  The graph of this function is rather simple, as we saw in Exercise 3.2.4.

b) Where does your symbolic answer not make sense?

c) Can you sketch the graph?

d) Why does the symbolic answer not work at the bad point?

part a

a) Differentiate y = (Sqrt[(x^2)+ 2 x + 1]) using the Chain Rule, the Power Rule, and the Superposition Rule.  The graph of this function is rather simple, as we saw in Exercise 3.2.4.

Solution of part a

part b

b) Where does your symbolic answer not make sense?

Solution of part b

The derivative of y = (Sqrt[(x^2)+ 2 x + 1]) is (dy/dx)(( =  2 + 2 x)/(2 (Sqrt[1 + 2 x + (x^2)]))).  This is undefined when x = -1 because when  x= -1 the denominator  is 0 and we cannot divide by 0.

part c

c) Can you sketch the graph?

Solution of part c

The graph of the function looks like this:

The graph of the derivative looks like this:

part d

d) Why does the symbolic answer not work at the bad point?

Solution of part d

When x = -1, the denominator is zero so the fraction is undefined making the function undefined at that point.

Exercise 3 : Three functions

Use the Superposition Rule and the Product Rule repeatedly to show in general that if f[x], g[x], and h[x] are smooth on an interval, then so are their sum and product and

((d (f[x]  +  g[x]  +  h[x]))/dx) = (df/dx)[x]  +  (dg/dx)[x] + (dh/dx)[x]

d[f[x] g[x] g[x]]/dx = (df/dx)[x] * g[x] * h[x] + f[x] * (dg/dx)[x] * h[x] + f[x] * g[x] * (dh/dx)[x]

(HINT: Let G[x] = g[x] * h[x] and apply the Product Rule to f[x]*G[x])

Solution

We know by the Superposition Rule that the derivative of the sum is the sum of the derivatives.

Let F[x] = f[x] + g[x], then dF[x]/dx = (df/dx)[x]  +  (dg/dx)[x]

And ((d(f[x]  +  g[x]  +  h[x]))/dx) = ((d (F[x]  +  h[x]))/dx) = dF[x]/dx + (dh/dx)[x]

= (df/dx)[x]  +  (dg/dx)[x] + (dh/dx)[x]

For the product let G[x] = g[x]*h[x].  And let K[x] = f[x]*G[x]. The Product Rule gives us:

(dK/dx) = (df/dx)[x] * G[x] + f[x] * (dG/dx)[x]

Now what is (dG/dx)[x]?  It is (dg/dx)[x] * h[x] + g[x] * (dh/dx)[x].  Substitute this and we get:

(dK/dx) = (df/dx)[x] * g[x] * h[x] + f[x] *(( dg)/dx)[x] * h[x] + f[x] * g[x] * (dh/dx)[x]

Exercise 4: Second Derivative Product Rule

When we compute the derivative of a function f[x] we get a new function f'[x].  If we think of this as a function, we could differentiate again: Let g[x] = f'[x] and compute g'[x].  The derivative of g[x] is called the second derivative of f[x] and is denoted g'[x] = f''[x] or

General Form			Example

y = f[x]				y = f[x] = (x^3)

(dy/(dx ))= f'[x]			(dy/(dx )) = f'[x] = 3 (x^2)

(((d^2) y)/(dx^2)) = f''[x]			(((d^2) y)/(dx^2)) = f''[x] = 3 * 2 x = 6 x

a) Differenciate the general Product Rule identity to get a formula for

(((d^2)(f[x] * g[x]))/(dx^2))

b) Let h[x] = f[x] * g[x] and use your rule to compute (((d^2)(h))/(dx^2))[1] if f[1] = 2, (df/dx)[1] = 3, (((d^2)(f))/(dx^2))[1] = 5, g[1] = -3, (dg/dx)[1] = 4. and (((d^2)(g))/(dx^2))[1] = -2.

Part a

a) Differenciate the general Product Rule identity to get a formula for

(((d^2)(f[x] * g[x]))/(dx^2))

Solution to a

The general Product Rule identity is ((d(f[x] * g[x]))/dx) = df[x]/dx*g[x] + f[x]dg[x]/dx.

We see here that we have two products added together. We know by the Superposition Rule that  the derviative of [df[x]/dx*g[x] + f[x]dg[x]/dx] is the derivative of each product added together .  To find the derivative of each product we need the Product Rule.  Both the Superposition Rule and the Product Rule are combined to get:

D(((d(f[x] * g[x]))/dx)) = D(df[x]/dx *g[x] + f[x]dg[x]/dx)

In other words, D(((d(f[x] * g[x]))/dx)) = [Product Rule] + [Product Rule]

Which can be simplified as:

(((d^2)(f[x] * g[x]))/(dx^2)) = (((d^2)(f[x] ))/(dx^2))*g[x] + (df[x]/dx*dg[x]/dx) + (df[x]/dx*dg[x]/dx)+ f[x]*(((d^2)(g[x]))/(dx^2))
	= (((d^2)(f[x] ))/(dx^2))*g[x]  + 2(df[x]/dx*dg[x]/dx)+ f[x]*(((d^2)(g[x]))/(dx^2))

Part b

b) Let h[x] = f[x] * g[x] and use your rule to compute (((d^2)(h))/(dx^2))[1] if f[1] = 2, (df/dx)[1] = 3, (((d^2)(f))/(dx^2))[1] = 5, g[1] = -3, (dg/dx)[1] = 4. and (((d^2)(g))/(dx^2))[1] = -2.

Solution to b

From part a we know

(((d^2)(f[x] * g[x]))/(dx^2)) = (((d^2)(f[x] ))/(dx^2))*g[x] + 2(df[x]/dx*dg[x]/dx)+ f[x]*(((d^2)(g[x]))/(dx^2))

Let h[x]= f[x]*g[x].  Thus we have:

(((d^2)(h))/(dx^2)) = (((d^2)(f[x] ))/(dx^2))*g[x] + 2(df[x]/dx*dg[x]/dx)+ f[x]*(((d^2)(g[x]))/(dx^2))

and
		(((d^2)(h))/(dx^2))[1] = (((d^2)(f))/(dx^2))[1] * g[1] + 2((df/dx)[1]*(dg/dx)[1]) + f[1]*(((d^2)(g))/(dx^2))[1]

We substitute to get:

(((d^2)(h))/(dx^2))[1]  =  5*(-3) + 2(3 * 4) + 2*(-2)   	=  -15 + 24 - 4     = 5

Therefore (((d^2)(h))/(dx^2))[1] = 5.