Implicit differentiation

Implicit differentiation involves differentiating implicit functions. An implicit function is an implicit relation between variables. Differentiating an implicit function leads to differentiate the independent variable with respect to the dependent variable. It's basically finding the derivative using the notation dy/dx.

Two methods can be used:

1) You explicit the function

Example 1

Find the derivative of 3xy = 2
Let's explicit the function: y = 2/3x
Let's calculate the derivative: dy/dx = d/dx(2/3x)
                                                          = -2(3x)'/(3x)²
                                                          = -6./9x²
                                                          = -2/3x²
2) If expliciting is not possible, you make transformations in order to find the derivative.

The rules and formulas used to calculate the derivative of different forms of functions apply in the calculations of the derivative of an implicit function.

Since an implicit function is a relationship between the independent and the dependent variable the the application of the derivative rules might seem odd. Let's familiarize ourselves with the derivatives of some expressions where the derivative rules are applied.

Example 2.   Let's y be a function of x find the derivative of y³ with respect to x.
Let's u = y³. we have two functions: u and y. U is a function of y and y is a function of x. U is a composite function. The chain rule has to be applied in order to find the derivative. The formula to apply here is du/dx = du/dy.dy/dx
du/dx = d(y³)/dy.dy/dx
           = 3y²dy/dx.
Example 3 Find the derivative of u = 2x²y
du/dx = d(2x²y)
Let's apply the constant rule
du/dx = 2d(x²y)
 Let's apply the product rule:
du/dx = 2[d/dx(x²)y+x² dy/dx]
          = :2(2xy+x²dy/dx)
du/dx = 4xy+2x²dy/dx 
Example 4 Find the derivative of 3y³+x²y = x-3
Let's differentiate both sides:
d/dx(3y³+x²y) = d/dx(x-3)
3y²dy/dx+2xy+x²dy/dx = 1
3y²dy/dx+x²dy/dx = 1-2xy
(3y²+x²)dy/dx = 1-2xy
dy/dx = 1-2xy/3y²+x²

Practice. Find the derivatives of the implicit functions:
1) x²+y² = 15
2) 3y²-siny = x²
3) x²+2xy-y = 2
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Derivative of exponential functions

Derivative of f(x) = b^x 

In the expression above b is a positive real number and is called the base of the exponential function.
The formula to calculate the derivative is d/dx[f(x)] = lnb.b<sup>x.

Rule: The derivative of an exponential function is equal to the product of the natural logarithm of the base by the function.</sup>

Example 1 calculate the derivative of f(x) = 2^x

The given function has the form f(x) = b^x. By applying the formula d/dx[f(x)] = lnb. b^x  d/dx[f(x)] = ln2.2^x

Derivative of f(x) = b^u 

Since f is a composite function where u is a function of x the derivative of f is d/dx [f(x)] = d/du(b^u).du/dx= lnb.b^u .u'

Rule: The derivative of an exponential function with base b is equal to the product of the natural logarithm of the base by the derivative of u.

Example 2. Calculate the derivative of f(x) = 3^2x. 

Let’s apply the formula for the derivative of f(x) = b^u which is d/dx[f(x)] = lnb.b^u.u’

d/dx[f(x)] = ln3.3^2x(2^x)’

                = ln3.3^2x.2

                = 2ln3.3^2x

Derivative of f(x) = e^x

The derivative f(x) = e^x is a special case of f(x) = b^x where b = e
Let's substitute b in the formula d/dx[f(x)] = lnb.b^x

d/dx[f(x)] = lne.e^x

Since lne = 1 d/dx[f(x)] = e^x

Rule: The derivative of the function f(x) = e^x is the function e^x itself. 

Derivative of f(x) = e^u 

Since f is a composite function where u is a function of x its derivative is given by the derivative of a composite function.
Then d/dx[f(x)] = d/du(e^u).du/dx = e^u.u’

Rule: The derivative of the composite exponential function with base e is equal to the product of the composite function by the derivative of the function u.

Example 3. Calculate the derivative of f(x) = e3x² ( Note this is not e.3x² but e with the exponent 3x²)

Let’s apply the formula for the derivative of f(x) = e^u which is d/dx[f(x)] = e^u.u’

d/dx[f(x)] = e3x².(3x²)’ 

                = e3x²(6x)

                = 6xe3x²

Summary

The derivative of f(x) = b^x where b>0 is d/dx(b^x) - lnb.b^x

The derivative of the composite function f(x) = b^u where u is a function of x is d/dx(b^u) = lnb.b^u u’

The derivative of f(x) = e^x is d/dx(ex) = e^x

 The derivative of f(x) = e^u is d/dx(e^u) = e^u.u’

Practice

Calculate the derivative of the following functions: 3x²

1) f(x) = e^6x 

2) f(x) = e3x²-4x+3 ( 3x²-4x+3 is the exponent )

3) f(x) = e^x-e^-x/e^x-e^-x

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Derivative of logarithmic functions

In this post I'll show some techniques to remember the formulas for logarithmic  functions. I'll do some examples and leave some exercises to practice.

Derivative of logarithmic functions 

Derivative of log_bx

d/dx (log_bx) = 1/xlnb
To remember this formula let's apply the following technique:
1) Multiply the number of which we calculate the logarithm by the natural logarithm of the base. The number here is x and the base is b. Therefore we have xlnb
2) Take the inverse of this product. The inverse of the product is 1/xlnx

Derivative of lnx

d/dx(lnx) = 1/x
The derivative of the logarithm of any number is equal to the inverse of this number.

Derivative of log_bu

Since log_bu is a composite function its derivative is given by d/dx(log_bu) = d/du(log_bu).du/dx

d/dx(log_bu) = 1/ulnnb.du/dx

Rule: The derivative of the logarithm of a composite function is equal to its derivative with respect to the new variable (u) multiplied by the derivative of the new variable (u) with respect to x.

Derivative of lnu 

Since u is a composite function we have d/dx(lnu) = d/du(lnu).du/dx
                                                                              = i/u.du/dx

Rule: The derivative of the natural logarithm of a composite function u is equal to the inverse of the function multiplied by its derivative with respect to x

Example 1, Calculate the derivative of y = x³log₅2x

The derivative of y is y" = (x³log₅2x)'

Let's apply the product rule:
Y' = (x³)'(log₅2x) + x³(log₅2x)'
The derivative of x³ is obvious. Let's calculate the derivative  of log₅2x
Let's write u = 2x we have (log₅u)' = d/du(log₅u),du/dx
                                                     = i/uln5.u'
                                                    = 1/2x.ln5.(2x)'
                                                   = 1/2x.ln5.2
                                                   = 1/xln5
let's go back to the derivative of y we have:
y' = 3x²log₅2x + x³.1/xlnx
   = 3x²log₅2x+x²/lnx

Example 2.  Calculate the derivative of y = ln(2x²-4x+3)

Let's write u = 2x²-4x+3
We have y = lnu
Then dy/dx = d/dx(lnu)
Since lnu is a composite function then dy/dx = d/du(lnu).du/dx
                                                                      = 1/u(4x-4)
Substitute u: dy/dx = (1/2x³-4x+3).(4x-4)
dy/dx = 4x-4/2x²-4x+3
           = 4(x-1)/2x³-4x+3

Practice

Calculate the derivative of the following functions;
1.log₅(2x+5)
2. 5/log(x+4)
3. ln(sinx)

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