Differentiation

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Differentiation rules

Now, that we understand the concept of a derivative, we will learn useful rules for evaluating derivatives more quickly and easily than by using the limit definition.

Discussion

Find the derivative of f(x)=x2 using the limit definition.
Find the derivative of g(x)=x3 using the limit definition.
What do you notice about the derivative of x raised to whole number powers?

Part (a)

Using the limit definition:

f′(x)=h→0limhf(x+h)−f(x)=h→0limh(x+h)2−x2=h→0limhx2+2xh+h2−x2=h→0limh2xh+h2=h→0lim(2x+h)=2x

Therefore, f′(x)=2x.

Part (b)

Using the limit definition:

g′(x)=h→0limhg(x+h)−g(x)=h→0limh(x+h)3−x3=h→0limhx3+3x2h+3xh2+h3−x3=h→0limh3x2h+3xh2+h3=h→0lim(3x2+3xh+h2)=3x2

Therefore, g′(x)=3x2

Part (c)

From parts (a) and (b) we have

dxdx2=2x,dxdx3=3x2

In each case the original exponent appears as a factor and the new exponent is one less. Hence for any whole number n,

dxdxn=nxn−1

In general, to find the derivative of xn (n∈N) starting with the limit definition of derivatives:

f′(x)=h→0limhf(x+h)−f(x)=h→0limh(x+h)n−xn

Use the factorisation of a difference of nth powers:

(x+h)n−xn=((x+h)−x)((x+h)n−1+(x+h)n−2x+⋯+xn−1)=h((x+h)n−1+(x+h)n−2x+⋯+xn−1)

Hence

h(x+h)n−xn=(x+h)n−1+(x+h)n−2x+⋯+xn−1

Now let h→0. Each term (x+h)kxn−1−k tends to xn−1, and there are n such terms. Therefore

f′(x)=nxn−1

for any whole number n. After learning more derivative rules, we can extend this rule to all real n.

Derivative of xⁿ where n is an integer

f(x)=xn, n∈Z⇒f′(x)=nxn−1📖

Why integer here ^ james q -- we have whole number but it works for all reals?

Checkpoint

Find the derivative of x7.

Select the correct option

Discussion

Using the limit definition, find the derivative of 3x2. What do you notice?
Using the limit definition, find the derivative of x2+x. What do you notice?

Part (a)

Let f(x)=3x2. By the limit definition

f′(x)=h→0limhf(x+h)−f(x)=h→0limh3(x+h)2−3x2

Expand and simplify:

f′(x)=h→0limh3(x2+2xh+h2)−3x2=h→0limh3x2+6xh+3h2−3x2=h→0limh6xh+3h2=h→0lim(6x+3h)=6x

We notice 6x=3⋅(2x), i.e. the derivative of 3x2 is three times the derivative of x2.

Part (b)

Let f(x)=x2+x. By the limit definition,

f′(x)=h→0limhf(x+h)−f(x)=h→0limh((x+h)2+(x+h))−(x2+x)

f′(x)=h→0limhx2+2xh+h2+x+h−x2−x=h→0limh2xh+h2+h=h→0lim(2x+h+1)=2x+1

We notice that 2x+1 is just the sum of the derivatives of x2 and of x; in other words, the derivative of a sum is the sum of the derivatives.

Derivatives of sums and scalar multiples

dxd(af(x))=af′(x)🚫

dxd(f(x)+g(x))=f′(x)+g′(x)🚫

dxd(af(x)+bg(x))=af′(x)+bg′(x)🚫

Checkpoint

Find the derivative of 2x3+4x2−6x+12.

Select the correct option

Exercise

Find dxd(10x3−x46).

Select the correct option

Derivative of xⁿ where n is rational

f(x)=xn, n∈Q⇒f′(x)=nxn−1📖

Examples:

f(x)=x4/3⇒f′(x)=34x1/3

Checkpoint

Find the derivative of x5/2.

Select the correct option

Discussion

Can you apply the power rule to expressions like (2x+1)2? Try applying the power rule over the whole expression. Then, try expanding first and applying the power rule to the resulting polynomial. Compare your answers.

First, apply the power rule to (2x+1)2 as if it were u2 with u=2x+1.

dxd((2x+1)2)=2(2x+1)=4x+2

Next, expand first and then differentiate term by term.

(2x+1)2=4x2+4x+1

dxd(4x2+4x+1)=8x+4

Since 4x+2=8x+4, the power rule does not work here.

For larger powers, expansion will become more difficult. For instance, we have no quick way to take the derivative of (2x+1)6. Notice that (2x+1)6 is the composition of the functions f(x)=2x+1 and g(x)=x6, and we can derive a rule for the derivative of compositions of functions called the chain rule.

Suppose we want to take the derivative of g(f(x)). First, write the derivative using the limit definition at a point:

(g(f(c)))′=x→climc−xg(f(c))−g(f(x)).

Then, we can multiply the numerator and denominator by f(c)−f(x) and split fractions, yielding

(g(f(c)))′=x→climf(c)−f(x)g(f(c))−g(f(x))⋅x→climc−xf(c)−f(x).

Now, we have two limits of fractions that can be rewritten as derivatives:

Notice that

x→climf(c)−f(x)g(f(c))−g(f(x))

is the definition of g′(c) except with f(c) and f(x) replacing c and x. Thus, this expression gives g′(f(c)). Then,

x→climc−xf(c)−f(x)

is exactly the definition of f′(c), so we have

(g(f(c)))′=g′(f(c))⋅f′(c).

Chain rule

(g(f(x)))′=g′(f(x))⋅f′(x)🚫

y=g(u) where u=f(x)

dxdy=dudg⋅dxdu📖

Checkpoint

Calculate the derivative of (2x+1)6.

Select the correct option

Discussion

Now, we find the derivative of ex together.

Write the limit definition of the derivative of ex. Then, factor ex out of the expression.

The resulting limit can be evaluated using the definition of ex:

ex=1+x+2!x2+…

Expand eh and evaluate h→0limheh−1.
Hence, state an expression for dxd(ex).

Part (a)

We use the limit definition with f(x)=ex.

f′(x)=h→0limhex+h−ex

Factor out ex:

f′(x)=h→0limhex(eh−1)=exh→0limheh−1

Part (b)

Expanding eh gives

eh=1+h+2!h2+⋯

Hence

h→0limheh−1=h→0limh(1+h+h2/2!+⋯)−1=h→0limhh+h2/2!+⋯=h→0lim(1+2!h+⋯)=1

Part (c)

From parts (a) and (b) we have

dxdex=exh→0limheh−1=ex⋅1

Hence

dxd(ex)=ex.

Derivative of e^x

f(x)=ex⇒f′(x)=ex📖

Checkpoint

Solve dxd(e2x+1).

Select the correct option

Discussion

We want to find the derivative of y=lnx.

By letting ey=x, find dydx.
Invert dydx to find dxdy, then simplify your expression in terms of x.

Part (a)

From y=lnx we have x=ey. Differentiate with respect to y:

dydx=ey.

Part (b)

Since

dydx=ey

we have

dxdy=ey1

and because ey=x this simplifies to

dxdy=x1.

Derivative of ln

f(x)=lnx⇒f′(x)=x1📖

Checkpoint

Find dxd(ln(7x−3)).

Select the correct option

By the limit definition,

(fg)′(c)=x→climc−xf(c)g(c)−f(x)g(x)

Add and subtract f(c)g(x) in the numerator:

(fg)′(c)=x→climc−xf(c)g(c)−f(c)g(x)+f(c)g(x)−f(x)g(x).

Then, by grouping and factoring we have

(fg)′(c)=x→climc−xf(c)(g(c)−g(x))+g(x)(f(c)−f(x)).

Now, we split fractions and limits:

(fg)′(c)=x→climf(c)c−x(g(c)−g(x))+x→climg(x)c−x(f(c)−f(x)).

Thus, because x→climf(c)=f(c) and x→climg(x)=g(c),

(fg)′(c)=f(c)(x→climc−x(g(c)−g(x)))+g(c)(x→climc−x(f(c)−f(x))).

By the definitions of f′ and g′,

(fg)′(c)=f′(c)g(c)+f(c)g′(c).

This is called the product rule.

Discussion

It is given that f(x)=3x2 and g(x)=x4.

Find (fg)′(x) by applying the product rule.
Confirm your answer to part (a) by multiplying f and g then using power rule to find the derivative of the product.

Part (a)

First compute the derivatives of f and g:

f′(x)=dxd(3x2)=6x,g′(x)=dxd(x4)=4x3

By the product rule,

(fg)′(x)=f′(x)g(x)+f(x)g′(x)=6x⋅x4+3x2⋅4x3=6x5+12x5=18x5.

Part (b)

Multiply the two functions first:

(fg)(x)=3x2⋅x4=3x2+4=3x6

Now apply the power rule:

dxd(3x6)=3⋅6x6−1=18x5

This agrees with the result from part (a).

Discussion

Now that we have the product and chain rules, you can derive the quotient rule.

Find an expression for (gf)′.

Write

gf=fg−1

By the product rule,

(gf)′=(fg−1)′=f′g−1+f(g−1)′

By the chain rule,

(g−1)′=−1⋅g−2g′=−g−2g′

Hence

(gf)′=f′g−1−fg−2g′.

This is a sufficient answer, but we can remove negative exponents and combine resulting fractions for a cleaner result:

f′g−1−fg−2g′=gf′−g2fg′=g2f′g−fg′.

Product and Quotient rule

The product and quotient

(uv)′=u′v+v′u📖

(vu)′=v2u′v−v′u📖

A popular shorthand for remembering the quotient rule is Lo2LoDHi−HiDLo.

Checkpoint

Find the derivative of x2ex.

Select the correct option

Animation?

Derivatives of sin and cos

f(x)=sinx⇒f′(x)=cosx📖

g(x)=cosx⇒g′(x)=−sinx📖

Exercise

Given that f(x)=sin(x2−1)ex, find f′(x).

Select the correct option

Exercise

Find dxd[cos(sinx⋅lnx)].

Select the correct option

Discussion

Using the derivatives of sinx and cosx, find the derivative of tanx.

By the quotient rule dxd(u/v)=v2vu′−uv′ with u=sinx, v=cosx:

dxdtanx=cos2xcosx⋅dxdsinx−sinx⋅dxdcosx=cos2xcosx⋅cosx−sinx⋅(−sinx)=cos2xcos2x+sin2x=cos2x1(using sin2x+cos2x=1)=sec2x

Therefore,

dxdtanx=sec2x.

Derivative of tan(x)

f(x)=tanx⇒f′(x)=sec2(x)📖

Checkpoint

Find the derivative of extanx.

Select the correct option

Discussion

Now that we know the derivatives for each of the trigonometric functions, we can find the derivatives of their reciprocals.

Find the derivative of secx. Simplify any fractions using tan or reciprocal functions.
Find the derivative of cosecx. Simplify any fractions using tan or reciprocal functions.
Find the derivative of cotx. Simplify any fractions using tan or reciprocal functions.

Part (a)

We write secx=(cosx)−1 and use the chain rule:

dxdsecx=dxd(cosx)−1=−1(cosx)−2⋅(−sinx)=cos2xsinx=cosx1⋅cosxsinx=secxtanx

Hence the derivative is

dxdsecx=secxtanx.

Part (b)

We write cosecx=(sinx)−1 and apply the chain rule:

dxdcosecx=dxd(sinx)−1=−1(sinx)−2⋅cosx=−sin2xcosx=−sinx1⋅sinxcosx=−cosecxcotx

Hence, the derivative is

dxdcosecx=−cosecxcotx.

Part (c)

We write cotx=(tanx)−1 and apply the chain rule:

dxdcotx=dxd(tanx)−1=−1(tanx)−2⋅dxdtanx=−tan2xsec2x=−sin2x/cos2x1/cos2x=−sin2x1=−cosec2x

Hence the derivative is

dxdcotx=−cosec2x.

Derivatives of reciprocal trig functions

f(x)=secx⇒f′(x)=secxtanx📖

g(x)=cosecx⇒g′(x)=−cosecxcotx📖

h(x)=cotx⇒h′(x)=−cosec2(x)📖

Discussion

We will walk through the proof for the derivative rule for arcsinx.

First, given y=arcsinx, how can we rewrite the equation so that it is only in terms of sin?
Hence, write an expression for dydx.
We want to write our final solution in terms of x, so rewrite cosy in terms of x. Recall that siny=x and use trigonometric identities.
Finally, take the reciprocal of dydx to get dxdy in terms of x.

Part (a)

Since arcsin is the inverse of sin (on [−2π,2π]), applying sin to both sides of

y=arcsinx

gives

siny=x.

Part (b)

Differentiate siny=x with respect to y:

dyd(siny)=dyd(x)⟹cosy=dydx.

Part (c)

Using sin2y+cos2y=1 and siny=x gives

cos2y=1−sin2y=1−x2.

Since y∈[−2π,2π] we have cosy≥0, so

cosy=√1−x2.

Part (d)

Since from parts (b) and (c) we have

dydx=cosy=√1−x2

taking reciprocals gives

dxdy=dydx1=√1−x21

We conclude that dxd(arcsinx)=√1−x21.

Following the same steps for arccosx yields dxd(arccosx)=√1−x2−1 since x=cosy⟹dydx=−siny.

Now, we prove the last rule for an inverse trigonometric function: arctanx.

If y=arctanx, then tany=x, so we have

dydx=sec2y.

Substituting with the trigonometric identity sec2y=tan2y+1,

dydx=tan2y+1=x2+1.

Therefore, we conclude that the derivative of y=arctanx is

dxdy=x2+11.

Derivatives of inverse trig functions

f(x)=arcsinx⇒f′(x)=√1−x21📖

g(x)=arccosx⇒g′(x)=√1−x2−1📖

h(x)=arctanx⇒h′(x)=1+x21📖

Checkpoint

Take the derivative of xarcsinx.

Select the correct option

Discussion

Utilize the derivative of ex to find the derivative of ax where a is any positive base.
Utilize the derivative of lnx to find the derivative of logax for any positive a.

Part (a)

Set y=ax. Since ax=exlna, we have

dxdy=dxd(exlna)=exlnadxd(xlna)=exlnalna=axlna

Hence dxdax=axlna.

Part (b)

Since logax=lnalnx with lna constant, we have

dxdlogax=dxd(lnalnx)=lna1dxd(lnx)=lna1⋅x1=xlna1

Hence

dxdlogax=xlna1

Derivatives of a^x and log_a(x)

f(x)=ax⇒f′(x)=axlna📖

g(x)=logax⇒f′(x)=xlna1📖

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