# Consider the Curve $y= x-x^7.$ How do I find the slope of the tangent line to the curve at the point $(1, 0).$

I have no idea how to solve this question.

a) Find the slope of the tangent line to the curve at the point $(1, 0)$.

Yet, how did they arrive at this answer using this formula:

$$m=\lim_{h\to0}\frac{f(a+h)-f(a)}h$$

Now the steps are shown as so:

(a) Using Definition 1: $m=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ with $f(x)=x-x^\color{red}7$ and $P(\color{red}1,\color{red}0)$,

\begin{align} m&=\lim_{x\to1}\frac{f(x)-\color{red}0}{x-\color{red}1}=\lim_{x\to1}\frac{x-x^\color{red}7}{x-\color{red}1}=\lim_{x\to1}\frac{x(1-x^\color{red}6)}{x-\color{red}1}\\&=\lim_{x\to1}\frac{x(1-x)\color{red}{(1+x+x^2+x^3+x^3+x^4+x^5)}}{x-\color{red}1}\\&=\lim_{x\to1}\left[-x\color{red}{\left(1+x+x^2+x^3+x^3+x^4+x^5\right)}\right]=-1(\color{red}6)=\color{red}{-6}. \end{align}

(b) An equation of the tangent line is \begin{align} &y-f(a)=f'(a)(x-a)\\\implies&y-f(\color{red}1)=f'(\color{red}1)(x-\color{red}1)\\\implies&y-\color{red}0=\color{red}{-6}(x-\color{red}1),\text{ or }y=-\color{red}{-6}x+\color{red}6 \end{align}

…but I don't understand them. Can someone explain these simply?

• i think answer should be -6 . – avz2611 Feb 22 '15 at 16:21
• Can you be more clear about exactly which of these steps you don't understand? – Gregory Grant Feb 22 '15 at 16:21
• The second line 4th equation. How was that expanded? – Cetshwayo Feb 22 '15 at 16:27
• $1-x^6 \text{ was factored } \\ \text{ You could divide } -x^6+1 \text{ by} -x+1 \text{ to find that other factor if you don't know the formula that is }$ – randomgirl Feb 22 '15 at 16:31

Take the derivative of $f(x)=x-x^7$ which is $f'(x)=1-7x^6$ and evaluate it in $x_0=1$: you'll obtain $f'(1)=1-7=-6$ which is the slope of the tangent line to the curve $y=x-x^7$ at the point $(1,0)$. I think there must be a typo in your book.
• In general it's dangerous to speak about "algorithm" in Mathematics. Think at integration: there isn't a unique way to solve them, you have to experience yourself. However in this case you have to know some basic matters in calculus. The question you asked was based on the fact that the definition of derivative of a function $f$ in a given point $x_0$ is the angular coefficient of the tangent line to the graphic of the curve defined from your function $f$ at the given point $(x_0,f(x_0))$. – Joe Feb 22 '15 at 18:16