Common Derivatives and Integrals Cheat Sheet, Cheat Sheet of Calculus

Download Common Derivatives and Integrals Cheat Sheet and more Cheat Sheet Calculus in PDF only on Docsity! Common Derivatives and Integrals Provided by the Academic Center for Excellence 1 Reviewed June 2008 Common Derivatives and Integrals Derivative Rules: 1. Constant Multiple Rule [ ] uccu dx d ′= , where c is a constant. 2. Sum and Difference Rule [ ] vuvu dx d ′±′=± 3. Product Rule [ ] uvvuuv dx d ′+′= 4. Quotient Rule 2 v vuuv v u dx d ′−′ =    5. Constant Rule, [ ] 0=c dx d 6. Power Rule [ ] unuu dx d nn ′= −1 7. Power Rule [ ] 1=x dx d 8. Derivative Involving Absolute Value [ ] ( ) 0, ≠′= uu u u u dx d 9. Derivative of the Natural Logarithmic Function [ ] u u u dx d ′ =ln 10. Derivative of Natural Exponential Function [ ] uee dx d uu ′= Provided by the Academic Center for Excellence 2 Common Derivatives and Integrals Example 1: Find the derivative of ( ) ( )( )22 2334 xxxxf +−= Since there are two polynomials multiplied by each other, apply the third derivative rule, the Product Rule, to the problem. This is the result of the Product Rule: Now, take the derivative of each term inside of the brackets. Multiple derivative rules are used, including the Sum and Difference Rule, Constant Rule, Constant Multiple Rule, and Power Rule. When applied, the result is: ( ) ( )( ) ( )( )xxxxxxf 64234034 22 −+++−=′ Simplify: ( ) ( )( ) ( )( )xxxxxxf 6423434 22 −++−=′ Multiply the polynomials by each other: ( ) ( ) ( )3232 12188121216 xxxxxxf −−++−=′ ( ) 3232 12188121216 xxxxxxf −−++−=′ Combine like terms to get a simplified answer: ( ) 12182424 23 +−+−=′ xxxxf Integral Formulas: Indefinite integrals have +C as an arbitrary constant. 1. ( ) ( )duufkduukf∫ ∫= , where k is a constant. 2. ( ) ( )[ ] ( ) ( )∫∫∫ ±=± duugduufduuguf 3. ∫ += Cudu ( ) ( ) [ ] ( ) [ ]2222 34232334 xx dx d xx dx d xxxf −+++−=′ Provided by the Academic Center for Excellence 5 Common Derivatives and Integrals Once the multiplication has been completed in the numerator of the fraction, the result is: ( ) ( ) ( )( )( )x xx xf 2 22 cos cossin + =′ Remember that ( ) ( ) 1cossin 22 =+ xx ; therefore, substitute 1 for ( ) ( )xx 22 cossin + in the answer. The final result is: ( ) ( )xxf 2cos 1 =′ ( ) ( )xxf 2sec=′ Integrals of Trigonometric Functions: 1. ( ) ( )∫ +−= Cuduu cossin 2. ( ) ( )∫ += Cuduu sincos 3 ( ) ( )∫ +−= Cuduu coslntan 4. ( ) ( )∫ += Cuduu sinlncot 5. ( ) ( ) ( )∫ ++= Cuuduu tanseclnsec 6. ( ) ( ) ( )∫ ++−= Cuuduu cotcsclncsc 7. ( ) ( )∫ += Cuduu tansec2 8. ( ) ( )∫ +−= Cuduu cotcsc2 9. ( ) ( ) ( )∫ += Cuduuu sectansec 10. ( ) ( ) ( )∫ +−= Cuduuu csccotcsc Provided by the Academic Center for Excellence 6 Common Derivatives and Integrals Example 4: Evaluate ( ) ( )( )∫ − dxxx tancos Normal integration formulas are often used in addition to trigonometric formulas when doing trigonometric integration. For example, in this problem use integration formula 2: ( ) ( )( )∫ − dxxx tancos = ( ) ( )∫ ∫− dxxdxx tancos With the two smaller integrals, use trigonometric integration formulas 2 and 3 to find the solution: = ( ) ( )( ) Cxx +−− coslnsin Simplify: = ( ) ( ) Cxx ++ coslnsin Special Differentiation Rules: Chain Rule: In certain situations, there may be a differentiable function of u, such as ( )ufy = , and ( )xgu = , where ( )xg is a differentiable function of x. If this is the case, then ( )( )xgfy = is a differentiable function of x. To take the derivative of the composite function ( )( )xgfy = , use the formula: dx du du dy dx dy ⋅= This formula can be rewritten as: ( )( )[ ] ( )( ) ( )xgxgfxgf dx d ′′= Example 5: Find dx dy of the function ( )243 xy += . Provided by the Academic Center for Excellence 7 Common Derivatives and Integrals Use the formula dx du du dy dx dy ⋅= to find this derivative. Now ( )43 x+ = u, and the result is the function 2uy = . Using the Chain Rule, we have ( )( ) ( )4334 38432 xxxx dx dy +=+= , where ( )432 x du dy += And 34x dx du = Special Integration Formulas: U-Substitution: Some integrals cannot be solved by using only the basic integration formulas. In some of these cases, one can use a process called u-substitution. This process helps simplify a problem before solving it. Example 6: Solve ( )∫ + dxx 2 1 32 Referring to the given integral formulas, there are none that are able to solve this integral in its current form. When one comes to an integral in a form like this, it may be possible to simplify the integral to a form that is solvable by the given formulas. For this integral, u- substitution may be used. : Using U- Substitution shows that: u = 2 + 3x dx dx du dx 3=      dxdu 3= ( ) 332 =′+= x dx du Provided by the Academic Center for Excellence 10 Common Derivatives and Integrals = ]31 3 1 2 3 1 3 10 2 5 3 4 xxx + +  The problem is now integrated completely. Plug in the upper and lower bounds into the problem and simplify: = ( ) ( ) ( ) ( ) ( ) ( )( )1103101 2 5 3 2 5 1 3 4 3 3 4 2233 −+      −+      − = ( ) ( ) ( ) ( ) ( ) ( )( )1103101 2 5 9 2 5 1 3 4 27 3 4 −+      −+      − = ( )1030 2 5 2 45 3 4 3 108 −+      −+      − = 20 2 40 3 104 ++ = 2020 3 104 ++ = 40 3 104 + =      + 3 3 40 3 104 = 3 120 3 104 + = 3 224 Integration by Parts: Some integrals can not be evaluated by using only the 16 basic integral formulas shown above. An example of an integral like this would be ∫ dxxe x . There are formulas to find ∫ xdx and ∫ dxe x , but we do not have a formula in our list that can find ∫ dxxe x . To evaluate an integral like this, use a method called “Integration by Parts”. This Provided by the Academic Center for Excellence 11 Common Derivatives and Integrals method is used to evaluate integrals where there are two separate functions of x contained in the integral, usually represented as u and v. An integral containing two separate functions of x usually follows the format ∫udv . The answer to this integral is ∫∫ −= vduuvudv . This formula can be used multiple times in a problem, depending on the values for u, v, du, and dv. Example 8: Find ∫ xdxx ln3 : With the two separate functions of x, 3x and xln , choose which function is needed to be u and dv. It is usually best to assign simple functions to be dv. For this problem, make u = xln and dv = 3x . Now find v and du. To find v, take the integral of dv. For du, take the derivative of u. Once this is done, the result is: xu ln= 4 4 1 xv = x du 1 = 3xdv = With the new values of our variables, plug them into the formula ∫∫ −= vduuvudv : ∫∫           −= dx x xxxxdxx 1 4 1 ln 4 1 ln 443 Simplify: = ∫− dxxxx 34 4 1 ln 4 1 After simplifying the integral, the simple integral is left and can be used to solve these formulas: = C x xx +      + − + 134 1 ln 4 1 13 4 = Cxxx +     − 44 4 1 4 1 ln 4 1 = Cxxx +− 44 16 1 ln 4 1 Provided by the Academic Center for Excellence 12 Common Derivatives and Integrals = Cxxx +−      44 16 1 ln 4 1 4 4 = Cxxx +− 44 16 1 ln 16 4 = ( ) C xx + − 16 1ln4 4