Calculus is Coming Along….

Hello, my math friends!

It’s been awhile since I’ve posted on here; life sort of gets in the way. I still try to make it a habit to work on the site every day, even if it’s writing a few words…. 😉

Calculus is coming along, with most sections in Differential Calculus and Integral Calculus completed. I’ve also written study guides for both Differential and Integral Calculus, since there are so many things to “memorize” in Calc, so it seems.

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Below is an example of one of the Calculus Charts I’ve created on the Chain Rule.

As always, please keep spreading the word about SheLovesMath.com and let me know how I make the site better!

Lisa 🙂

 Function Derivative Using Chain Rule Notes $$\displaystyle f\left( x \right)={{\left( {5x-1} \right)}^{8}}$$ \displaystyle \begin{align}{f}’\left( x \right)&=8{{\left( {\color{red}{{5x-1}}} \right)}^{7}}\cdot \color{red}{5}\\&=40{{\left( {5x-1} \right)}^{7}}\end{align} Since the $$\left( {5x-1} \right)$$ is the inner function, after using the Power Rule, we have to multiply by the derivative of that function, which is 5. $$\displaystyle f\left( x \right)={{\left( {{{x}^{4}}-1} \right)}^{3}}$$ \displaystyle \begin{align}{f}’\left( x \right)&=3{{\left( {\color{red}{{{{x}^{4}}-1}}} \right)}^{2}}\cdot \left( {\color{red}{{4{{x}^{3}}}}} \right)\\&=12{{x}^{3}}{{\left( {{{x}^{4}}-1} \right)}^{2}}\end{align} Since the $$\left( {{{x}^{4}}-1} \right)$$ is the inner function, after using the Power Rule, we have to multiply by the derivative of that function, which is $$4{{x}^{3}}$$. $$\displaystyle \begin{array}{l}g\left( x \right)=\sqrt[4]{{16-{{x}^{3}}}}\\g\left( x \right)={{\left( {16-{{x}^{3}}} \right)}^{{\frac{1}{4}}}}\end{array}$$ \displaystyle \begin{align}{l}{g}’\left( x \right)&=\frac{1}{4}{{\left( {\color{red}{{16-{{x}^{3}}}}} \right)}^{{-\frac{3}{4}}}}\cdot \left( {\color{red}{{-3{{x}^{2}}}}} \right)\\&=-\frac{{3{{x}^{2}}}}{{4{{{\left( {16-{{x}^{3}}} \right)}}^{{\frac{3}{4}}}}}}=-\frac{{3{{x}^{2}}}}{{4\,\sqrt[4]{{{{{\left( {16-{{x}^{3}}} \right)}}^{3}}}}}}\end{align} Since the $$\left( {16-{{x}^{3}}} \right)$$ is the inner function, after using the Power Rule, we have to multiply by the derivative of that function, which is $$-3{{x}^{2}}$$. $$\displaystyle \begin{array}{l}f\left( t \right)=t\sqrt{{3t-2}}\\f\left( t \right)=t{{\left( {3t-2} \right)}^{{\frac{1}{2}}}}\end{array}$$ \displaystyle \begin{align}{f}’\left( t \right)&=t\left( {\frac{1}{2}} \right){{\left( {\color{red}{{3t-2}}} \right)}^{{-\frac{1}{2}}}}\cdot \left( {\color{red}{3}} \right)+{{\left( {3t-2} \right)}^{{\frac{1}{2}}}}\cdot 1\\&=\frac{{3t}}{2}{{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}+{{\left( {3t-2} \right)}^{{\frac{1}{2}}}}\\&={{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}\left( {\frac{{3t}}{2}+3t-2} \right)\\&={{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}\left( {\frac{{3t}}{2}+\frac{{6t}}{2}-\frac{4}{2}} \right)\\&=\left( {\frac{{9t-4}}{{2\sqrt{{3t-2}}}}} \right)\end{align} Use the Product Rule, since we have $$t$$’s in both expressions. Since the $$\left( {3t-2} \right)$$ is the inner function, we have to multiply by the derivative of that function, which is 3. Note that we also took out the Greatest Common Factor (GCF) $${{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}$$, so we could simplify the expression. $$\displaystyle y=\cos \left( {4x} \right)$$ $$\displaystyle \begin{array}{l}{y}’=-\sin \left( {\color{red}{{4x}}} \right)\cdot \color{red}{4}\\=-4\sin \left( {4x} \right)\end{array}$$ Since the $$\left( {4x} \right)$$ is the inner function (the argument of $$\text{sin}$$), we have to take multiply by the derivative of that function, which is 4. $$\displaystyle g\left( x \right)=\cos \left( {\tan x} \right)$$ \displaystyle \begin{align}{g}’\left( x \right)&=-\sin \left( {\color{red}{{\tan x}}} \right)\cdot \color{red}{{{{{\sec }}^{2}}x}}\\&=-{{\sec }^{2}}x\cdot \sin \left( {\tan x} \right)\end{align} Since the $$\left( {\tan x} \right)$$ is the inner function (the argument of $$\text{cos}$$), we have to multiply by the derivative of that function, which is $$\displaystyle {{\sec }^{2}}x$$. $$\displaystyle \begin{array}{l}f\left( x \right)={{\sec }^{3}}\left( {\pi x} \right)\\f\left( x \right)={{\left[ {\sec \left( {\pi x} \right)} \right]}^{3}}\end{array}$$   (Note second way to write this) \displaystyle \begin{align}{f}’\left( x \right)&=3\,{{\color{red}{{\sec }}}^{2}}\left( {\color{blue}{{\pi x}}} \right)\cdot \left( {\color{red}{{\sec \left( {\color{blue}{{\pi x}}} \right)\tan \left( {\color{blue}{{\pi x}}} \right)}}} \right)\color{blue}{\pi }\\&=3\pi {{\sec }^{3}}\left( {\pi x} \right)\tan \left( {\pi x} \right)\end{align} This one’s a little tricky, since we have to use the Chain Rule twice. When we take the derivative of $${{\left[ {\sec \left( {\pi x} \right)} \right]}^{3}}$$, we have to multiply by the derivative of $$\sec \left( {\pi x} \right)$$, which is $$\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi$$ (again, we had to multiply by the derivative of $$\pi x$$, which is $$\pi$$). $$\displaystyle \begin{array}{l}f\left( \theta \right)=2{{\cot }^{2}}\left( {2\theta } \right)+\theta \\f\left( \theta \right)=2{{\left[ {\cot \left( {2\theta } \right)} \right]}^{2}}+\theta \end{array}$$ \displaystyle \begin{align}{f}’\left( \theta \right)=&4\,\color{red}{{\cot }}\left( {\color{blue}{{2\theta }}} \right)\cdot \color{red}{{-{{{\csc }}^{2}}\left( {\color{blue}{{2\theta }}} \right)}}\cdot \color{blue}{2}+1\\&=1-8{{\csc }^{2}}\left( {2\theta } \right)\cot \left( {2\theta } \right)\end{align} This is another one where we have to use the Chain Rule twice.
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