Derivatives Part 12017-08-05T23:29:00+00:00

Derivatives

Click on the ANSWER tab for general solutions and examples for each.

 

Basic Derivatives

1. \(\frac{d}{dx}{{u}^{n}}\,\,\)                            ANSWER
\(\begin{array}{l}\frac{d}{dx}\,\,{{u}^{n}}=(n){{u}^{n-1}}{u}’\\Ex1:\,\,\frac{d}{dx}4{{x}^{5}}-15{{x}^{3}}+{{x}^{\frac{3}{2}}}=20{{x}^{4}}-45{{x}^{2}}+\frac{3}{2}{{x}^{\frac{1}{2}}}\\Ex2:\,\,\frac{d}{dx}7{{x}^{-4}}-{{x}^{-3}}+10{{x}^{\frac{1}{2}}}=-28{{x}^{-5}}+3{{x}^{-4}}+5{{x}^{-\frac{1}{2}}}\\=-\frac{28}{{{x}^{5}}}+\frac{3}{{{x}^{4}}}+\frac{5}{{{x}^{\frac{1}{2}}}}\end{array}\) Proof 

2. \(\frac{d}{dx}\,\,sin\,u\)                        ANSWER
\(\frac{d}{dx}\,\,sin\,u=\cos \,u\,{u}’\)    
Proof
 

3. \(\frac{d}{dx}\,\,\cos \,u\)                       ANSWER
\(\frac{d}{dx}\,\,\cos \,u=-\sin \,u\,{u}’\)  
Proof
 

4. \(\frac{d}{dx}\,\,\frac{1}{u}\)                             ANSWER
\(\frac{d}{dx}\,\,\frac{1}{u}=-\frac{1}{{{u}^{2}}}\,{u}’\)

5. \(\frac{d}{dx}\,\,\tan \,u\)                        ANSWER
\(\frac{d}{dx}\,\,\tan \,u={{\sec }^{2}}u\,{u}’\)  
Proof 

6. \(\text{Product}\,\,\text{Rule:}\,\,\,\,\frac{d}{dx}\,\,f(x)\bullet g(x)\,\,or\,\,\frac{d}{dx}\,\,uv\)              ANSWER
\(\begin{array}{l}\frac{d}{dx}\,\,f(x)\bullet g(x)=\,\,\,\,\,{f}'(x)\bullet g(x)+f(x)\bullet {g}'(x)\,\,\,\,\\or\,\,\\\frac{d}{dx}\,\,uv={u}’v+u{v}’\\Ex1:\,\,\frac{d}{dx}8{{x}^{2}}\tan x=16x\tan x+8{{x}^{2}}{{\sec }^{2}}x\\Ex2:\,\,\frac{d}{dx}\sin x\cos x=\cos x\cos x+\sin x(-\sin x)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={{\cos }^{2}}x-{{\sin }^{2}}x\end{array}\) \(\begin{array}{l}Ex3:\,\,\frac{d}{dx}{{x}^{2}}\sqrt{x-5}=2x{{(x-5)}^{\frac{1}{2}}}+{{x}^{2}}\frac{1}{2}{{(x-5)}^{-\frac{1}{2}}}\\\,\,\,\,\,\,\,\,\,\,now\,\,factor:\,\,=x{{(x-5)}^{-\frac{1}{2}}}(2(x-5)+\frac{1}{2}x)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=x{{(x-5)}^{-\frac{1}{2}}}(2x-10+\frac{1}{2}x)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{x}{\sqrt{x-5}}(\frac{5}{2}x-10)\,\,\,\,\,or\,\,\,\,\frac{5x(x-25)}{2\sqrt{x-5}}\,\end{array}\)

7. \(\text{Quotient}\,\,\text{Rule:}\,\,\,\,\frac{d}{dx}\,\,\frac{hi}{lo}\,\,or\,\,\frac{d}{dx}\,\,\frac{f(x)}{g(x)}\)                        ANSWER
\(\begin{array}{l}\ Ex1:\,\,\frac{d}{dx}\frac{\sin x}{{{x}^{2}}}=\frac{{{x}^{2}}\cos x\,-\sin x\,(2x)}{{{({{x}^{2}})}^{2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{{x}^{2}}\cos x\,-2x\sin x\,}{{{x}^{4}}}=\frac{x\cos x\,-2\sin x\,}{{{x}^{3}}}\\Ex2:\,\,\frac{d}{dt}500(1\,\,\,+\,\,\frac{4t}{50+{{t}^{2}}})=\frac{d}{dt}(500\,\,+\,\,\frac{2000t}{50+{{t}^{2}}})\\=\frac{(50+{{t}^{2}})2000-2000t(2t)}{{{(50+{{t}^{2}})}^{2}}}\\\frac{100,000+2000{{t}^{2}}-4000{{t}^{2}}}{{{(50+{{t}^{2}})}^{2}}}=\frac{-2000{{t}^{2}}+100,000}{{{(50+{{t}^{2}})}^{2}}}\\=\frac{-2000({{t}^{2}}-50)}{{{(50+{{t}^{2}})}^{2}}}\\Ex3:\,\,\frac{d}{dx}\,\,\,\frac{\sin x}{1-\cos x}=\frac{(1-\cos x)\cos x-\sin x(\sin x)}{{{(1-\cos x)}^{2}}}\\=\frac{\cos x-{{\cos }^{2}}x-{{\sin }^{2}}x}{{{(1-\cos x)}^{2}}}\\=\frac{\cos x-({{\cos }^{2}}x+{{\sin }^{2}}x)}{{{(1-\cos x)}^{2}}}\,\,\,\,\,\,\,\,remember\,\,\,{{\sin }^{2}}x+{{\cos }^{2}}x=1\\=\frac{\cos x-1}{{{(1-\cos x)}^{2}}}=\frac{-(-\cos x+1)}{{{(1-\cos x)}^{2}}}=\frac{-1}{1-\cos x}\end{array}\)

8. \(\text{Chain}\,\,\text{Rule}:\,\,\frac{d}{dx}f(g(2x))\)                                       ANSWER
\(\frac{d}{dx}f(g(2x))={f}'(g(2x))\bullet {g}'(2x)\bullet 2\) \(\begin{array}{l}\text{Differentiate}\,\,\text{the}\,\,\,\text{following:}\\1.\,\,h(x)=10\sin (7x-9)\\{h}'(x)=10\cos (7x-9)\bullet 7\\=70\cos (7x-9)\\\\2.\,\,p(x)=\sqrt[3]{7{{x}^{2}}+5x-3}={{(7{{x}^{2}}+5x-3)}^{\frac{1}{3}}}\\{p}'(x)=\frac{1}{3}{{(7{{x}^{2}}+5x-3)}^{-\frac{2}{3}}}(14x+5)\\=\frac{(14x+5)}{3{{(7{{x}^{2}}+5x-3)}^{\frac{2}{3}}}}\\\\3.\,\,k(x)=10{{\cos }^{4}}(7x-9)\\{k}'(x)=40{{\cos }^{3}}(7x-9)\bullet (-\sin (7x-9)\bullet 7)\\=-280{{\cos }^{3}}(7x-9)\sin (7x-9)\\\\4.\,\,\text{Product}\,\,\text{and}\,\,\text{chain}\\f(x)=4{{x}^{3}}{{(5x-8)}^{\frac{3}{2}}}\\{f}'(x)=12{{x}^{2}}{{(5x-8)}^{\frac{3}{2}}}+4{{x}^{3}}(\frac{3}{2}){{(5x-8)}^{\frac{1}{2}}}(5)\\=12{{x}^{2}}{{(5x-8)}^{\frac{3}{2}}}+30{{x}^{3}}{{(5x-8)}^{\frac{1}{2}}}\\=6{{x}^{2}}{{(5x-8)}^{\frac{1}{2}}}[2(5x-8)+5x]\\=6{{x}^{2}}{{(5x-8)}^{\frac{1}{2}}}(15x-16)\\\\5.\,\,h(x)=10f(g(8x))\\{h}'(x)=10{f}'(g(8x))\bullet {g}'(8x)\bullet 8\\=80{f}'(g(8x))\bullet {g}'(8x)\\\\6.\,\,\,y={{\sin }^{2}}(4x){{\cos }^{3}}(5x)\\{y}’=2\sin (4x)\cos (4x)(4)\bullet {{\cos }^{3}}(4x)+{{\sin }^{2}}(4x)3{{\cos }^{2}}(5x)\bullet (-\sin (5x))\bullet (5)\\=8\sin (4x)\cos (4x){{\cos }^{3}}(4x)-15{{\sin }^{2}}(4x){{\cos }^{2}}(5x)\sin (5x)\\\end{array}\) 

 

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Trigonometric Functions

1. \(\frac{d}{dx}\sin u\)                ANSWER
\(\begin{array}{l}\frac{d}{dx}\sin u\,\,=\cos u\bullet {u}’\\Ex:\,\,\frac{d}{dx}\sin (7x-12)\,\,=\cos (7x-12)\bullet 7\\=7\cos (7x-12)\end{array}\)

2. \(\frac{d}{dx}\cos u\)               ANSWER
\(\begin{array}{l}\frac{d}{dx}\cos u\,\,=-\sin u\bullet {u}’\\Ex:\,\,\frac{d}{dx}\cos ({{x}^{2}}+2x-3)\,\,=-\sin ({{x}^{2}}+2x-3)\bullet (2x+2)\\=-(2x+2)\sin ({{x}^{2}}+2x-3)\end{array}\)

3. \(\frac{d}{dx}\tan \,u\)               ANSWER
\(\begin{array}{l}\frac{d}{dx}\tan \,u\,={{\sec }^{2}}u\bullet {u}’\\Ex:\,\,\frac{d}{dx}\tan (5{{x}^{2}}-7x)\,\,={{\sec }^{2}}(5{{x}^{2}}-7x)\bullet (10x-7)\\=(10x-7){{\sec }^{2}}(5{{x}^{2}}-7x)\end{array}\)

4. \(\frac{d}{dx}\sec \,u\)               ANSWER
\(\begin{array}{l}\frac{d}{dx}\sec \,u\,=\sec u\tan u\bullet {u}’\\Ex:\,\,\frac{d}{dx}\sec ({{x}^{2}}-12)\,\,=\sec ({{x}^{2}}-12)\bullet \tan ({{x}^{2}}-12)\bullet 2x\\=2x\sec ({{x}^{2}}-12)\bullet \tan ({{x}^{2}}-12)\end{array}\)

5. \(\frac{d}{dx}\csc \,u\)               ANSWER
\(\begin{array}{l}Ex:\,\,\frac{d}{dx}\csc (\sqrt{x}-12)\,\,=-\csc (\sqrt{x}-12)\bullet \cot (\sqrt{x}-12)\bullet \frac{1}{2}{{x}^{-\frac{1}{2}}}\\=-\frac{1}{2}{{x}^{-\frac{1}{2}}}\csc (\sqrt{x}-12)\bullet \cot (\sqrt{x}-12)\end{array}\)

6. \(\frac{d}{dx}\cot \,u\)               ANSWER
\(\begin{array}{l}\frac{d}{dx}\cot \,u\,=-{{\csc }^{2}}u\bullet {u}’\\Ex:\,\,\frac{d}{dx}\cot ({{x}^{2}}+{{x}^{\frac{1}{3}}})\,\,=-{{\csc }^{2}}({{x}^{2}}+{{x}^{\frac{1}{3}}})\bullet (2x+\frac{1}{3}{{x}^{-\frac{2}{3}}})\\=-(2x+\frac{1}{3}{{x}^{-\frac{2}{3}}}){{\csc }^{2}}({{x}^{2}}+{{x}^{\frac{1}{3}}})\end{array}\)

 

 

Inverse Trigonometric Functions

1. \(\frac{d}{dx}\arcsin u\)                ANSWER
\(\begin{array}{l}\frac{d}{dx}\arcsin u=\frac{du}{\sqrt{1-{{u}^{2}}}}\,\,\,and\,\,\,\,\,\frac{d}{dx}\arccos u=\frac{-du}{\sqrt{1-{{u}^{2}}}}\\Ex:\frac{d}{dx}\arcsin (7x-1)=\frac{7}{\sqrt{1-{{(7x-1)}^{2}}}}\end{array}\)

2. \(\frac{d}{dx}\arctan u\)               ANSWER
\(\begin{array}{l}\frac{d}{dx}\arctan u=\frac{du}{1+{{u}^{2}}}\,\,\,and\,\,\,\,\,\frac{d}{dx}\text{arccot}\,u=\frac{-du}{1+{{u}^{2}}}\\Ex:\frac{d}{dx}\arctan ({{x}^{2}}-x)=\frac{2x-1}{1+{{({{x}^{2}}-x)}^{2}}}\end{array}\)

3. \(\frac{d}{dx}\text{arcsec}\,u\)               ANSWER
\(\begin{array}{l}\frac{d}{dx}\text{arcsec}\,u=\frac{du}{\left| u \right|\sqrt{{{u}^{2}}-1}}\,\,\,and\,\,\,\,\,\frac{d}{dx}\text{arccsc}\,u=\frac{-du}{\left| u \right|\sqrt{{{u}^{2}}-1}}\\Ex:\frac{d}{dx}\text{arcsec }({{x}^{3}})=\frac{3{{x}^{2}}}{\left| {{x}^{3}} \right|\sqrt{{{x}^{6}}-1}}\end{array}\)

 

Exponential Functions

1. \(\frac{d}{dx}{{e}^{u}}\)                ANSWER
\(\begin{array}{l}\frac{d}{dx}{{e}^{u}}={{e}^{u}}\,{u}’\\Ex1:\,\,\frac{d}{dx}{{e}^{4x-{{x}^{2}}}}=(4-2x){{e}^{4x-{{x}^{2}}}}\\Ex2:\,\,\frac{d}{dx}{{e}^{\sin (8x)}}=8\sin (8x){{e}^{\sin (8x)}}\\Ex3:\,\,\frac{d}{dx}\arctan (3x){{e}^{-{{x}^{2}}}}\\=\frac{3}{1+9{{x}^{2}}}{{e}^{-{{x}^{2}}}}+\arctan (3x)\bullet (-2x){{e}^{-{{x}^{2}}}}\\={{e}^{-{{x}^{2}}}}\left( \frac{3}{1+9{{x}^{2}}}-2x\arctan (3x) \right)\\\end{array}\)

2. \(\frac{d}{dx}{{a}^{u}}\)                ANSWER
\(\begin{array}{l}\frac{d}{dx}{{a}^{u}}=\ln a\bullet {{a}^{u}}\,{u}’\\Ex1:\,\,\frac{d}{dx}{{4}^{{{x}^{3}}}}=\ln 4\bullet {{4}^{{{x}^{3}}}}\bullet 3{{x}^{2}}=3{{x}^{2}}\ln 4\bullet {{4}^{{{x}^{3}}}}\\Ex2:\,\,\frac{d}{dx}{{7}^{-k{{x}^{2}}}}=\ln 7\bullet {{7}^{-k{{x}^{2}}}}\bullet -2kx\\=-2kx\ln 7\bullet {{7}^{-k{{x}^{2}}}}\,\,\text{(where}\,\,\text{k}\,\,\text{is}\,\,\text{a}\,\,\text{constant)}\\Ex3:\,\,\frac{d}{dx}{{\sin }^{3}}(2x){{k}^{-2x}}\\=3{{\sin }^{2}}(2x)\cos (2x)\bullet 2\bullet {{k}^{-2x}}+{{\sin }^{3}}(2x)\bullet \ln k\bullet {{k}^{-2x}}(-2)\\=6{{\sin }^{2}}(2x)\cos (2x)\bullet {{k}^{-2x}}-2{{\sin }^{3}}(2x)\ln k\bullet {{k}^{-2x}}\\=2{{\sin }^{2}}(2x){{k}^{-2x}}(3\cos (2x)-\sin (2x)\ln k)\\\,\text{(where}\,\,\text{k}\,\,\text{is}\,\,\text{a}\,\,\text{constant)}\\\end{array}\)

 

Logarithmic Functions

1. \(\frac{d}{dx}\ln u\)                ANSWER
\(\begin{array}{l}\frac{d}{dx}\ln u=\frac{{{u}’}}{u}\\Ex1:\,\,\frac{d}{dx}\ln (8-{{x}^{2}})=\frac{-2x}{(8-{{x}^{2}})}\\Ex2:\,\,\frac{d}{dx}\ln (\arctan (4x))=\frac{\frac{4}{1+16{{x}^{2}}}}{\arctan (4x)}\\=\frac{4}{(1+16{{x}^{2}})\arctan (4x)}\\Ex3:\,\,\frac{d}{dx}4{{x}^{2}}\ln (x-1)=8x\ln (x-1)+\frac{4{{x}^{2}}(1)}{(x-1)}\\=8x\ln (x-1)+\frac{4{{x}^{2}}}{(x-1)}\\\end{array}\)

2. \(\frac{d}{dx}{{\log }_{a}}u\)             ANSWER
\(\begin{array}{l}\frac{d}{dx}{{\log }_{a}}u=\text{(Change}\,\,\text{of}\,\,\text{base}\,\,\text{first}\,\,’\text{a }\!\!’\!\!\text{ }\,\,\text{is}\,\,\text{a}\,\,\text{constant)}\\=\frac{d}{dx}\frac{\ln u}{\ln a}=\frac{{{u}’}}{\ln a\,\,\bullet u}\,\,\text{remember}\,\,\ln a\,\,\text{is}\,\,\text{just}\,\,\text{a}\,\,\text{constant}\\Ex1:\,\,\frac{d}{dx}{{\log }_{3}}(8-{{x}^{2}})=\frac{d}{dx}\frac{\ln (8-{{x}^{2}})}{\ln 3}\\=\frac{-2x}{\ln 3(8-{{x}^{2}})}\\Ex2:\,\,\frac{d}{dx}{{\log }_{8}}({{x}^{2}}-3x+12)=\frac{d}{dx}\frac{\ln ({{x}^{2}}-3x+12)}{\ln 8}\\=\frac{2x-3}{\ln 8({{x}^{2}}-3x+12)}\\Ex3:\,\,\frac{d}{dx}{{x}^{2}}{{\log }_{2}}(x-1)=\frac{d}{dx}{{x}^{2}}\frac{\ln (x-1)}{\ln 2}\\=2x\frac{\ln (x-1)}{\ln 2}+{{x}^{2}}\frac{1}{(x-1)\ln 2}\\=\frac{x}{\ln 2}\left[ 2\ln (x-1)+\frac{x}{(x-1)} \right]\\\end{array}\)

 

 

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