Difficulty finding the slope of the tangent line with two variables inside the equation This was a question posted in my lecture that me and my friends are unable to solve. The professor said this should be done and learned in high school, but here I am in university unable to complete this question. It might have something to do with differentiation, but I still am unable to figure out how to complete this question. Any help regarding it would be greatly appreciated.
Calculate the slope of the tangent to:
$f(x)=(x^2+1)^q$
when $q = 3$, and $x = -1$.
 A: If $f(x):=(x^2+1)^3$ then $$f’(x)=3(x^2+1)^2\cdot 2x \implies f’(-1)=6\cdot 2^2\cdot (-1)=-24.$$ Hence the slope of the tangent line to $y=f(x)$ when $x=-1$ is $m=-24$.
A: The statement "calculate the slope of the line tangent to $f(x)=(x^2+1)^q$ when $q = 3$ and $x = -1$" means that you first need to replace $q$ in the function expression with the one given to you as part of the problem to obtain the actual function and then find the slope of the line tangent to the graph of this function in general and finally your task is to calculate the slope of this line at a point $x=-1$.
The first order of business here is that we need to find the first derivative of this function. The first derivate of a function gives us the slope of the tangent line at any point $x$ (that's the reason why it's sometimes called the slope function). For that, we're going to use simple basic differentiation rules plus the famous chain rule:
$$
f'(x)=\left((x^2+1)^3\right)'=3(x^2+1)^2(x^2+1)'=3(x^2+1)^2\cdot 2x=
6x(x^2+1)
$$
We've now got our slope function. Let's find out what it is equal to at the point $x=-1$ by simply plugging in the given number into our slope function:
$$
f'(-1)=6\cdot(-1)\cdot((-1)^2+1)^2=-6\cdot 2^2=-24
$$
And $-24$ is the answer to the question.
