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${\begin{array}{lcl}(a^{2}+b^{2})^{3}&\neq &a^{6}+b^{6}\\(a^{2}+b^{2})^{3}&=&(a^{2}+b^{2})(a^{2}+b^{2})(a^{2}+b^{2})\\&=&(a^{4}+2a^{2}b^{2}+b^{4})(a^{2}+b^{2})\\&=&(a^{6}+2a^{4}b^{2}+a^{2}b^{4}+a^{4}b^{2}+2a^{2}b^{4}+b^{6})\end{array}}$

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63. Suppose that L is the tangent line at $x=x_{0}$ to the graph of the cubic equation $y=ax^{3}+bx$. Find the x-coordinate of the point where L intersects the graph a second time.

First, we find the derivative of y:

$y'=3ax^{2}+b\,$

This gives us the slope of L at $x_{0}$. Next we designate the point $(x_{0},y_{0})$ as the point where L is tangent to y(Note: $x_{0}$ is treated as a constant from here on out).

Now, using the point-slope form of a line, we define L:

$L=y-y_{0}=(3ax_{0}^{2}+b)(x-x_{0})$

We can write $y_{0}$ in terms of $x_{0}$ using the original equation:

$y_{0}=ax_{0}^{3}+bx_{0}$

Then,

${\begin{array}{lcl}L=y-(ax_{0}^{3}+bx_{0})&=&(3ax_{0}^{2}+b)(x-x_{0})\\y-ax_{0}^{3}-bx_{0}&=&3ax_{0}^{2}x-3ax_{0}^{3}+bx-bx_{0}\\y&=&3ax_{0}^{2}x-2ax_{0}^{3}+bx\end{array}}$

Now that we have the above formula for the tangent line L, we set it equal to the original cubic equation and find all the solutions:

${\begin{array}{lcl}ax^{3}+bx&=&3ax_{0}^{2}x-2ax_{0}^{3}+bx\\ax^{3}&=&3ax_{0}^{2}x-2ax_{0}^{3}\\x^{3}&=&3x_{0}^{2}x-2x_{0}^{3}\\x^{3}-3x_{0}^{2}x+2x_{0}^{3}&=&0\end{array}}$

To factor the above we will use synthetic division. We already know that $x-x_{0}$ is a factor, because $x_{0}$ is where L is tangent to the above.

$x-x_{0}{\overline {\vert x^{3}+0x^{2}-3x_{0}^{2}x+2x_{0}^{3}}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&&&\\&|&1&&&\\\end{matrix}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&x_{0}&&\\&|&1&&&\\\end{matrix}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&x_{0}&&\\&|&1&x_{0}&&\\\end{matrix}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&x_{0}&x_{0}^{2}&\\&|&1&x_{0}&&\\\end{matrix}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&x_{0}&x_{0}^{2}&\\&|&1&x_{0}&-2x_{0}^{2}&\\\end{matrix}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&x_{0}&x_{0}^{2}&-2x_{0}^{3}\\&|&1&x_{0}&-2x_{0}^{2}&\\\end{matrix}}$

${\begin{matrix}x_{0}&|&1&0&-3x_{0}^{2}&2x_{0}^{3}\\&|&&x_{0}&x_{0}^{2}&-2x_{0}^{3}\\&|&1&x_{0}&-2x_{0}^{2}&0\\\end{matrix}}$

Since the remainder is 0, this confirms that $x-x_{0}$ is a factor.

${\begin{array}{lcl}(x^{2}+x_{0}x-2x_{0}^{2})(x-x_{0})&=&0\\(x-x_{0})(x+2x_{0})(x-x_{0})&=&0\\(x-x_{0})^{2}(x+2x_{0})&=&0\\\end{array}}$

Thus, L crosses $y=ax^{3}+bx\,$ at $x=-2x_{0}\,$

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Given:

${\begin{aligned}f'(x)&=\lim _{h\to 0}{f(x+h)-f(x) \over (h)}\\f(x)&=x^{n}\end{aligned}}$

Prove:

$f'(x)=nx^{n-1}\,$

${\begin{aligned}f'(x)&=\lim _{h\to 0}{(x+h)^{n}-x^{n} \over {h}}\\&=\lim _{h\to 0}{(x^{n}+nx^{n-1}h+{\cfrac {n(n-1)}{2!}}x^{n-2}h^{2}+\cdots +h^{n})-x^{n} \over {h}}\\&=\lim _{h\to 0}{nx^{n-1}h+{\cfrac {n(n-1)}{2!}}x^{n-2}h^{2}+\cdots +h^{n}) \over {h}}\\&=\lim _{h\to 0}{h(nx^{n-1}+{\cfrac {n(n-1)}{2!}}x^{n-2}h+\cdots +h^{n-1}) \over {h}}\\&=\lim _{h\to 0}{nx^{n-1}+{\cfrac {n(n-1)}{2!}}x^{n-2}h+\cdots +h^{n-1}}\\&=\lim _{h\to 0}{nx^{n-1}}\\&=nx^{n-1}\end{aligned}}$

Given:

$f(x)=c\,$

Where c is a constant, prove:

$f'(x)=0\,$

${\begin{aligned}f'(x)&=\lim _{h\to 0}{f(x+h)-f(x) \over {h}}\\&=\lim _{h\to 0}{c-c \over {h}}\\&=\lim _{h\to 0}{0}\\&=0\end{aligned}}$

Given:

$f(x)=mx+b\,$

Prove:

$f'(x)=m\,$

${\begin{aligned}f'(x)&=\lim _{h\to 0}{f(x+h)-f(x) \over {h}}\\&=\lim _{h\to 0}{(m(x+h)+b)-(mx+b) \over {h}}\\&=\lim _{h\to 0}{mx+mh+b-mx-b \over {h}}\\&=\lim _{h\to 0}{mh \over {h}}\\&=\lim _{h\to 0}{m}\\&=m\end{aligned}}$