The derivative of e with a constant compared with a constant and an unknown

Question

My understanding when taking the derivative;

$$e^x = e^x$$ $$e^{10} = e^{10}$$

But happens when there is an arithmetic symbol in the power?

$$e^{10 -t}$$

I've seen this answer given "$(-1)e^{10 - t}$"

Question 1: Why do we not deduct $1$ from $t$? example;

$$(-1)e^{10 - t - 1}$$

Question 2: Am I correct to assume that the first figure in the power (in respect only to $e$) DOES NOT do anything? No deducting by $1$ and no multiplying in front of the $e$?

Question 3 What happens with second values if they are not an unknown? Are they the only ones that multiply by $e$? Do they get deducted by $1$? Example;

$$e^{10 - 5} = (-5)e^{10 - 4}$$

Answer 1

The derivative of $e^{10}$ is $0$ since $e^{10}$ is a constant. But if $$y(t)=e^{10-t}$$

notice $y(t)$ is a composite function, i.e. $$y(t)=y(x(t))=e^{10-t}$$ where $x(t)=10-t$. Thus, we must invoke the Chain Rule to differentiate it. Namely, $$y(t)=y(x(t))\implies\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{d\left(e^{\overbrace{10-t}^{x(t)}} \right)}{dx}\cdot\frac{d\left(10-t\right)}{dt}$$ $$=e^{10-t}\cdot(-1)=-e^{10-t}$$

Answer 2

The derivative of a constant is always zero. so (e^C)'=0$ where $C$ is a constant.

For $e^{10-t}$ the genral rule is $$(e^{f(t)})'=e^{f(t)}f'(t)$$ $$\implies (e^{10-t})'=(e^{10-t})(10-t)'=-e^{10-t}$$

Answer 3

@Anon First, your statement that $e^{10}$ is not correct. It is a constant and the derivative of a constant is 0. Of course, if you meant $e^x$, then yes, its derivative is $e^x$.

Question 1: You're thinking of derivatives of monomials, where for example, d/dx of $x^3$ is 3$\cdot x^{3-1}=3\cdot x^2$. However, the Exponential Function IS NOT a monomial. It is a constant raised to some function of x.

Question 2: Not sure what your question is here, but I suspect that better understanding of derivatives and the Chain Rule will resolve it.

Question 3: your statement is not correct. $e^{10-4}$ is $e^6$ and this is a constant. The derivative of a constant is 0.

To address your comment "didn't realise that e to the power of something with arithmetic operations was part of the chain rule"; it comes into play because a variable substitution mandated the Chain Rule.

To see what I mean, another way to look at taking the derivative of $y=e^{10-t}$ is to apply the derivative operator to both sides of the equation:

$\frac {d}{dt}$ [y] = $\frac {d}{dt}$ [$e^{10-t}$]

By definition, the LHS becomes y'. Before continuing on the RHS, let $\mu$ = $10-t$

Our equation is now y' = $\frac {d}{dt}$ [$e^{\mu}$]. Whoa! Take notice that we're applying d/dt to a function of $\mu$, and that's why the Chain Rule comes into play. The Chain Rule effectively says you can change d/dt to d/d$\mu$ ONLY IF you multiply by d$\mu$/dt. Thus,

y' = $\left( \frac {d}{d\mu} [e^{\mu}]\right) \cdot\left(\frac {d\mu}{dt}\right)$

The first part of the RHS is simply $e^{\mu}$. So now we have

y' = $e^{\mu} \cdot\left(\frac {d\mu}{dt}\right)$

So now we must determine d$\mu$/dt. We do this by referring to our variable substitution equation, and applying the Derivative Operator to both sides:

$\frac {d}{dt} [\mu]$ = $\frac {d}{dt}$ [10-t}] Again by definition, the LHS becomes d$\mu$/dt. So we have

$\frac {d\mu}{dt}$= $\frac {d}{dt}$ [10-t}] = $\frac {d}{dt}$ [10] - $\frac {d}{dt}$ [t]

The derivative of the constant, 10, by definition, is 0. The derivative of t is 1. Therefore,

$\frac {d\mu}{dt}$= (0)-(1)=-1. Substituting yields

y' = $e^{\mu} \cdot (-1)$. And re-substituting the definition of $\mu$ yields

y' = -$e^{10-t}$