Differentiation 3

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YouTube lecture recording from October 2020

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Exponentials and Partial Differentiation

Examples of applying chain rule to the exponential function

1. $\displaystyle y=e^{-ax}$

   • Let $\displaystyle u=-ax\Rightarrow\frac{{\rm d}u}{{\rm d}x}=-a$.

   • Thus $\displaystyle y=e^u$ and

   • $\displaystyle \frac{{\rm d}y}{{\rm d}u}=e^u~~\Rightarrow~~\frac{{\rm d}y}{{\rm d}x}=\frac{{\rm d}y}{{\rm d}u}\times\frac{{\rm d}u}{{\rm d}x}=e^u\times(-1.=-ae^{-ax}.$

2. $\displaystyle y = e^{x^2}$

   • Then, letting $\displaystyle u = x^2$

   • $\displaystyle \frac{{\rm d}}{{\rm d}x}e^{x^2}=\frac{{\rm d}y}{{\rm d}x}=\frac{{\rm d}y}{{\rm d}u}\times\frac{{\rm d}u}{{\rm d}x}=e^u\cdot 2x = e^{x^2}\cdot 2x.$

$\displaystyle \frac{{\rm d}}{{\rm d}x}e^{f(x)}=e^{f(x)}f'(x)$ for any function $f(x)$

Example with the natural logarithm

1. $\displaystyle y=\ln(a-x)^2=2\ln(a-x)=2\ln u$.

   • Let $\displaystyle u=(a-x)$:

   • $\displaystyle \Rightarrow {{\rm d}u\over {\rm d}x}=-1~~{\rm and~~~~~}{{\rm d}y\over {\rm d}u}={2\over u}~~~{\rm Thus~~~~}{{\rm d}y\over {\rm d}x}={2\over u}\times (-1)={-2\over a-x}$

$\displaystyle \frac{{\rm d}}{{\rm d}x}\ln(f(x)) = {f'(x)\over f(x)}$

The Derivative of $a^x$

$\displaystyle \frac{{\rm d}}{{\rm d}x}a^x = \frac{{\rm d}}{{\rm d}x}e^{\left({x\cdot\ln a}\right)} = e^{\left({x\cdot\ln a}\right)}\frac{{\rm d}}{{\rm d}x}{\left({x\cdot\ln a}\right)} =a^x\cdot\ln a$

$\displaystyle \frac{{\rm d}}{{\rm d}x}a^{f(x)} = a^{f(x)}\cdot \ln a\cdot f'(x)$

Sympy Example

Copy
import sympy as sp
x, a = sp.symbols('x a') # declare the variables x and a
f = sp.Function('f')     # declare a function dependent on another variable
sp.diff(a**f(x),x)       # write the expression we wish to evaluate

$\displaystyle a^{f{\left(x \right)}} \log{\left(a \right)} \frac{d}{d x} f{\left(x \right)}$

The Derivative of $\displaystyle \log_a x\,\,$

$\displaystyle \frac{{\rm d}}{{\rm d}x}\log_a x = \frac{{\rm d}}{{\rm d}x}\left({1\over{\ln a}}\cdot\ln x\right) = \left({1\over{\ln a}}\right)\cdot\frac{{\rm d}}{{\rm d}x}\ln x = \left({1\over{\ln a}}\right)\cdot{1\over {x}} = {1\over{x\cdot\ln a}}$

$\displaystyle \frac{{\rm d}}{{\rm d}x}\log_a f(x) = {{f'(x)} \over {f(x){(\ln 1.}}}$

Sympy Example

Copy
x, a = sp.symbols('x a')  # declare the variables x and a
f = sp.Function('f')      # declare a function dependent on another variable
sp.diff(sp.log(f(x),a),x) # write the expression we wish to evaluate

$\displaystyle \frac{\frac{d}{d x} f{\left(x \right)}}{f{\left(x \right)} \log{\left(a \right)}}$

Further examples

1. Product Rule: Let $\displaystyle y = x^2\,e^x$. Then:

   $\displaystyle {{dy\over dx}}={d\over dx}x^2e^x={d\over dx}x^2\cdot e^x+x^2\cdot{d\over dx}e^x = (2x + x^2)e^x$

2. Quotient Rule: Let $\displaystyle y = {{e^x}\over x}$. Then:

   $\displaystyle {{dy\over dx}}={{{{d\over dx}e^x}\cdot x - e^x\cdot {d\over dx}x}\over {x^2}}={{e^x\cdot x - e^x\cdot 1\over {x^2}}}={{x - 1}\over x^2}e^x$

3. Chain Rule: $\displaystyle y = e^{x^2}$. Then, letting $\displaystyle f(x) = x^2$:

   $\displaystyle \frac{{\rm d}}{{\rm d}x}e^{x^2} = e^{f(x)}f'(x) = e^{x^2}\cdot 2x$

4. $\displaystyle y=\ln (x^2 + 1)$. Then, letting $f(x) = x^2+1$:

   $\displaystyle \frac{{\rm d}}{{\rm d}x}\ln(x^2+1) = {f'(x)\over f(x)} = {2x\over {x^2+1}}$

5. $\displaystyle {{\rm d}\over {\rm d}x}2^{x^3}=2^{x^3}\cdot\ln 2\cdot 3x^2$

6. $\displaystyle {{\rm d}\over {\rm d}x}10^{x^2+1}=10^{x^2+1}\cdot\ln 10\cdot 2x$

7. $\displaystyle \frac{{\rm d}}{{\rm d}x}\log_{10}(7x+5)={7\over {(7x+5)\cdot \ln10}}$

8. $\displaystyle \frac{{\rm d}}{{\rm d}x}\log_2(3^x+x^4)={{3^x\cdot\ln3 + 4x^3}\over{\ln 2\cdot(3^x+x^4)}}$

Functions of several variables: Partial Differentiation

$\displaystyle \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}f(x,y) = f_x(x,y)$

Example 1

$\displaystyle f(x,y)=z=x^2-2y^2$ > $\displaystyle f_x={\partial z\over \partial x}=2x\qquad\rm{and}\qquad f_y={\partial z\over \partial y}=-4y$

Example 2

$\displaystyle \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}\,\left(3x^2y+5xy^2\right)$ > $\displaystyle \qquad =3y\cdot 2x + 5y^2\cdot 1$ > $\displaystyle \qquad =6xy+5y^2$

$\displaystyle \frac{\partial z}{\partial y}=\frac{\partial}{\partial y}\,\left(3x^2y+5xy^2\right)\,$ > $\displaystyle \qquad =3x^2\cdot 1 + 5x\cdot 2y = 3x^2+10xy$

Sympy example

Copy
x, y = sp.symbols('x y')
sp.diff(3*x**2*y + 5*x*y**2,x)

Higher-Order Partial Derivatives

1. With respect to $x$ twice:

   $\displaystyle \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right) =\frac{\partial^2z}{\partial x^2} =z_{xx}$

2. With respect to $y$ twice:

   $\displaystyle \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right) =\frac{\partial^2z}{\partial y^2} =z_{yy}$

3. First with respect to $x$, then with respect to $y$:

   $\displaystyle \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right) =\frac{\partial^2z}{\partial y\partial x} =z_{xy}$

4. First with respect to $y$, then with respect to $x$:

   $\displaystyle \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right) =\frac{\partial^2z}{\partial x\partial y} =z_{yx}$

Example: LaPlace's equation for equilibrium temperature distribution on a copper plate

$\displaystyle T_{xx} + T_{yy} = 0$

$\displaystyle T_x(x,y)=0-2x=-2x\qquad\rm{so}\qquad T_{xx}(x,y)=-2$

$\displaystyle T_y(x,y)=2y-0=2y\qquad\rm{so}\qquad T_{yy}(x,y)=2$

$\displaystyle T_{xx}(x,y)+T_{yy}(x,y) = 2 + (-2) = 0$

$\displaystyle z_x = 2xy - y^2$ > $\displaystyle z_{xx}=2y$

$\displaystyle z_y = x^2 - 2xy$ > $\displaystyle z_{yy} =-2x$

$\displaystyle z_{xx}+z_{yy}=2y-2x\ne 0$

Copy
T1 = y**2 - x**2
sp.diff(T1, x, x) + sp.diff(T1, y, y)

Copy
T2 = x**2*y - x*y**2
sp.diff(T2, x, x) + sp.diff(T2, y, y)

A Note on the Mixed Partials $f_{xy}$ and $f_{yx}$

Example

$\displaystyle z_{xy}=\frac{\partial}{\partial y}\left(2xy^3+6x\right)=6xy^2$

$\displaystyle z_{yx}=\frac{\partial}{\partial x}\left(3x^2y^2-8y^3\right)=6xy^2$

$\displaystyle {\rm~i.e.~~~~}~{\partial\over \partial x}\biggr({\partial z\over \partial y}\biggl)~~~~={\partial\over \partial y}\biggr({\partial z\over \partial x}\biggl)$

Introductory problems

Main problems

Extension problems