Substitution Rule

Author: John J Weber III, PhD Corresponding Textbook Sections:

Prerequisite Knowledge

Calculus I Rules

FunctionDerivativeFunctionAntiderivative
f(x)f(x)f(x)F(x)
f(x)=a; aRf(x)=0f(x)=a; aRF(x)=ax+C
f(x)=axn;
a,nR
f(x)=naxn1f(x)=axn;
a,nR, n1
F(x)=xn+1n+1+C
f(x)=ln(x)f(x)=1xf(x)=1x=x1F(x)=ln(x)+C
f(x)=logb(x);
b>0,b1, x>0
f(x)=1xln(b)f(x)=1xln(b);
b>0,b1
F(x)=(1ln(b))ln(x)+C
f(x)=ax;
a>0,a1
f(x)=ln(a)axf(x)=ax;
a>0,a1
F(x)=axln(a)+C
f(x)=exf(x)=exf(x)=exF(x)=ex+C
f(x)=sin(x)f(x)=cos(x)f(x)=cos(x)F(x)=sin(x)+C
f(x)=cos(x)f(x)=sin(x)f(x)=sin(x)F(x)=cos(x)+C
f(x)=tan(x)f(x)=sec2(x)f(x)=sec2(x)F(x)=tan(x)+C
f(x)=cot(x)f(x)=csc2(x)f(x)=csc2(x)F(x)=cot(x)+C
f(x)=sec(x)f(x)=sec(x)tan(x)f(x)=sec(x)tan(x)F(x)=sec(x)+C
f(x)=csc(x)f(x)=csc(x)cot(x)f(x)=csc(x)cot(x)F(x)=csc(x)+C
f(x)=sin1(x)f(x)=11x2f(x)=11x2F(x)=sin1(x)+C
f(x)=cos1(x)f(x)=11x2f(x)=11x2F(x)=cos1(x)+C
f(x)=tan1(x)f(x)=11+x2f(x)=11+x2F(x)=tan1(x)+C
f(x)=cot1(x)f(x)=11+x2f(x)=11+x2F(x)=cot1(x)+C
f(x)=sec1(x)f(x)=1xx21f(x)=1xx21F(x)=sec1(x)+C
f(x)=csc1(x)f(x)=1xx21f(x)=1xx21F(x)=csc1(x)+C

Check Your Work

Use Technology to Verify a Derivative

Mathematica

NOTE: On all Assessments, Mathematica may be used to verify derivatives. You are expected to be able to show how to differentiate any function in the course.

Warnings:

  1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.
  2. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.
  3. Mathematica may return a result in a different, but equivalent, form than your answer. In these cases, use FTC-I to check your work.

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Created: Wednesday, 22 December 2021 - 15:50 (EDT)