Mistakes regarding derivatives

Regarding concepts

A function can be continuous and non-differentiable at the same time. Can we say that if the function is not differentiable at a point this is caused by a division by zero? No. This idea comes from the fact that we first see the derivative in a relationship with the tangent line. We have this limit: [math]\displaystyle{ \lim_{h \ \to \ 0} \frac{f(x + h) - f(x)}{h} }[/math] and [math]\displaystyle{ h \to 0 }[/math] means that we are making the distance between two points in the number line close to zero. Are we dividing by zero? No and this is the level of abstraction that we are faced with when we first learn about limits. Sometimes we naively assume that to say that a limit doesn't exist is the same as to say that we tried to divide by zero.

[math]\displaystyle{ \frac{d}{dx} \sin(x^2) \neq \cos(2x) }[/math]. This silly mistake is a result of misunderstanding the chain rule. Better say, from trying to memorize it without understanding it. The chain rule says that [math]\displaystyle{ f(g(x))' = g'(x)f'(g(x)) }[/math] and not [math]\displaystyle{ f(g(x))' = f'(g'(x)) }[/math]. The correct answer is: [math]\displaystyle{ 2x \cos(x^2) }[/math]

[math]\displaystyle{ \frac{\partial}{\partial x} \sin(x + y) \neq \cos(x) }[/math].
[math]\displaystyle{ \frac{\partial}{\partial x} \sin(x + y) \neq \cos(y) }[/math].
[math]\displaystyle{ \frac{\partial}{\partial x} \sin(x + y) \neq \cos(y + 1) }[/math].
[math]\displaystyle{ \frac{\partial}{\partial x} \sin(x + y) \neq \cos(x + 1) }[/math]. The correct answer for this partial derivative is [math]\displaystyle{ \cos(x+y) }[/math]. This mistake is the same that happens with single variable functions, now making us do even worse when there are multiple variables.

[math]\displaystyle{ \frac{\partial}{\partial x} \sin(x + xy) = (y + 1) \cos(x+xy) }[/math] How about this? Now [math]\displaystyle{ y }[/math] may be a constant, but [math]\displaystyle{ xy }[/math] is no longer a constant. To calculate this derivative we cannot forget the chain rule! So x in respect to x is 1 and xy in respect to x is y. The rest is the same as the above example.

[math]\displaystyle{ f(x) = x^{\frac{1}{2}} }[/math] and [math]\displaystyle{ f(x) = \frac{1}{x} }[/math]. A very common mistake is to forget that roots are rational exponents and negative exponents mean the inverse of a number or variable. This leads to calculating derivatives the wrong way.

[math]\displaystyle{ f(x) = x^n \implies f'(x) = nx^{n \ - \ 1} }[/math]. A not uncommon misstep here is to do this [math]\displaystyle{ f'(x) = (n - 1)x^{n \ - \ 1} }[/math]. I'd say that among all errors in mathematics, any formula that has a [math]\displaystyle{ (n \pm 1) }[/math] term in it is susceptible to this very mistake. Another mistake is to forget the constant [math]\displaystyle{ f(x) = cx^n }[/math] if there is one (there is always a constant 1 by the way).

[math]\displaystyle{ f(x) = e^x \implies f'(x) = e^x }[/math]. Sometimes we memorize this and make this mistake: [math]\displaystyle{ f(x) = e^{2x} \not\!\!\!\implies f'(x) = e^{2x} }[/math] because me forget that this is a composite function.