6.2 Standard Derivatives
Example:
Differentiatey=x12.Solution: So using the rule that if y = xn then dy
dx =nxn−1. We have n = 1
2. This gives the answer:
dy dx = 1
2×x12−1=x−12 2 = 1
2√ x
So far we have only considered polynomials which have a coefficient of 1. But how do we differentiate y= 4x2where the coefficient is equal to 4? When we have a coefficientawe have the rule:
Ify=axn then dy
dx =a×n×xn−1
Note: The first rule that we discussed is just a special case of the above rule whena= 1.
So going back toy= 4x2 we can see that the coefficienta= 4 andxis raised to the power ofn= 2. So using the rule, ify=axn then dy
dx =a×n×xn−1. We have:
dy
dx = 4×2×x2−1= 8x1= 8x
IMPORTANT:When differentiating any functionf(x) multiplied by some constanta, findingdxd [af(x)], we can take the constantaout of the derivative so:
d dx
af(x)
=a d dx
f(x)
For example d
dx(2x3) = 2× d
dx(x3) = 2×3x2= 6x2.
Example:
Differentiatey= 4x−12Solution: This example combines everything we have looked at so far. So using the rule:
Ify=axn then dy
dx =a×n×xn−1 we havea= 4 and n=−1
2. This gives the answer:
dy
dx = 4× −1
2×x−12−1=−2x−32
Physics Example:
Coulomb’s Law states that the magnitude of the attractive forceF be- tween two charges decays according to the inverse square of the separating distancer:F = 1 r2
If we plot a graph ofF on they-axis againstr on thex-axis, what is the gradient?
Solution: To find the gradient we find the derivative with respect tordenoted dF dr of:
F = 1 r2 =r−2
All the examples so far have only containedxandyhowever in this question we have the variables dy
Below we have the rules for differentiating trigonometric functions:
Ify= sin(x) then dy
dx = cos(x)
Ify= cos(x) then dy
dx =−sin(x)
Ify= tan(x) then dy
dx = sec2(x)
When we have constants multiplying the trigonometric functions and coefficients multiplying our xwe can use the following rules:
Ify=asin(bx) then dy
dx =a×b×cos(bx)
Ify=acos(bx) then dy
dx =−a×b×sin(bx)
Ify=atan(bx) then dy
dx =a×b×sec2(bx) Note: These rules only work whenxis being measured in radians.
Example:
Differentiate:1. y= sin(9x) 2. y= 4 cosx
2 3. y=−2 tan (−7x)
Solution: Using the rules above:
1. Ify=asin(bx) then dy
dx =a×b×cos(bx)
Note that we havey= sin(9x) soa= 1 and b= 9. So dy
dx = 9 cos(9x) 2. Ify=acos(bx) then dy
dx =−a×b×sin(bx) Note that we havey= 4 cosx
2
soa= 4 andb= 1 2
dy
dx =−4× ×1
2×sinx 2
=−2 sinx 2
3. Ify=atan(bx) then dy
dx =a×b×sec2(bx)
Note that we havey=−2 tan (−7x) soa=−2 andb=−7 dy
dx = (−2)× −7 sec2(−7x)
=
−14 sec2(−7x) = 14 sec2(7x)
Physics Example:
In X-ray crystallography, the Bragg equation relates the distance d be- tween successive layers in a crystal, the wavelength of the X-raysλ, an integer nand the angle through which the X-rays are scatteredθin the equation:λ= 2d n sinθ What is the rate of change ofλwithθ?
Solution: The question is asking us to find dλ
dθ. So asnanddare constants we can use the rule:
Ify=asin(bx) then dy
dx =a×b×cos(bx) So we havea=
2d n
andb= 1 with λas y andθasxtherefore:
dλ dθ =
2d n
×1×cosθ= 2d n cosθ.
Differentiating Exponential Functions
The exponential function ex is special, since when we differentiateex, we just obtainex again. This is shown in the rule below:
Ify=exthen dy
dx =ex
We can also differentiate when thexhas a coefficient, for exampley=e4x, and also when the exponential is multiplied by a constant, for exampley= 3ex. We combine these into the rule below:
Ify=aebx then dy
dx =a×b×ebx
Example:
Differentiate the following with respect tox:1. y=e2x 2. y= 8 exp
−x 3
Solution:
1. Using the rule
Ify=aebx then dy
dx =a×b×ebx note y=e2x soa= 1 andb= 2 hence dy
dx = 2e2x 2. Using the rule
Ify=aebx then dy
dx =a×b×ebx −x
−1 dy
× −1
×
−x
−8 −x
Ify= ln(x) then dy
dx = 1 x
We can differentiate when thexhas a coefficient, for exampley= ln(4x), and also when the logarithmic function has a coefficient, for exampley= 3 ln(x). We combine these into the rule below:
Ify=aln(bx) then dy
dx = a x
Note: In the above rule, the b disappears and it forms a nice exercise in the practice of using the laws of logarithms hence we have included the algebra in the next example. We do have to differenti- ate a sum however so if you are unsure on that step then check out the next section for more information.
Example:
Fory=aln(bx) find dy dxSolution: Using laws of logarithms we can writey=aln(bx) as:
y=a(ln(x) + ln(b)) Now differentiating this as a sum gives:
dy dx = d
dx[a(ln(x) + ln(b))]
=⇒ dy dx =a d
dx[ln(x)] +a d dx[ln(b)]
Now as d
dx[ln(x)] = 1
x and d
dx[ln(b)] = 0 as ln(b) is a constant, we have:
dy dx = a
x
Note: This derivation is included as it provides an advanced example of the use of the laws of logarithms with differentiation.
Example:
Differentiate:1. y= ln(3x) 2. y=−4
5ln x
2
Solution:
1. We use the rule:
Ify=aln(bx) then dy dx = a
x Note that we havey= ln(3x) soa= 1 andb= 3. Hence dy
dx = 1 x 2. We use the rule:
Ify=aln(bx) then dy dx = a
x Note that we havey=−4
ln x
soa=−4
andb=1
. Hence dy
=−4
×1
=− 4
Differentiating Hyperbolic Functions
Below we have the rules for differentiating hyperbolic functions:
Ify= sinh(x) then dy
dx = cosh(x)
Ify= cosh(x) then dy
dx = sinh(x)
Ify= tanh(x) then dy
dx = sech2(x)
When we have constants multiplying the hyperbolic functions and coefficients multiplying ourxwe can use the following rules:
Ify=asinh(bx) then dy
dx =a×b×cosh(bx)
Ify=acosh(bx) then dy
dx =a×b×sinh(bx)
Ify=atanh(bx) then dy
dx =a×b×sech2(bx)
Note: that the derivative of cosh(x) is sinh(x) and NOT−sinh(x). We can prove these derivatives by differentiating the exponential forms of the hyperbolic functions:
Example:
Differentiate cosh(x) by converting to the exponential form.Solution: Begin by writing cosh(x) as
cosh(x) = ex 2 +e−x
2 Differentiating each term:
d
dxcosh(x) = ex 2 −e−x
2
= sinh(x)