Prove the formulas for differentiating a sum and a product: $$ d\left(\omega_{1}+\omega_{2}\right)=d \omega_{1}+d \omega_{2} . $$ and $$ d\left(\omega^{k} \wedge \omega^{\prime}\right)=d \omega^{k} \wedge \omega^{\prime}+(-1)^{k} \omega^{k} \wedge d \omega^{l} . $$

We want to prove that \(d(\omega_1+\omega_2)=d\omega_1+d\omega_2\). To do this, we will use the sum rule of differentiation. The sum rule states that the derivative of the sum of two functions is equal to the sum of their derivatives. Mathematically, this can be written as: $$ \frac{d}{dt}(f(t)+g(t))=\frac{df(t)}{dt}+\frac{dg(t)}{dt} $$ Applying the sum rule to our functions \(\omega_1\) and \(\omega_2\), we find that: $$ \frac{d}{dt}(\omega_1(t)+\omega_2(t))=\frac{d\omega_1(t)}{dt}+\frac{d\omega_2(t)}{dt} $$ This is equivalent to: $$ d(\omega_1+\omega_2)=d\omega_1+d\omega_2 $$ Thus, we have proved the first formula for differentiating a sum.

Now, we want to prove that when differentiating a product of two functions (\(\omega^k \wedge \omega^\prime\)), the derivative is given by the formula: $$ d(\omega^{k} \wedge \omega^{\prime})=d \omega^{k} \wedge \omega^{\prime}+(-1)^{k} \omega^{k} \wedge d \omega^{l} $$ To do this, we will use the product rule of differentiation. The product rule states that the derivative of the product of two functions is equal to the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function. Mathematically, this can be written as: $$ \frac{d}{dt}(f(t)g(t))=\frac{df(t)}{dt}g(t)+f(t)\frac{dg(t)}{dt} $$ Applying the product rule to our functions \(\omega^k\) and \(\omega^\prime\), we find that: $$ \frac{d}{dt}(\omega^k(t) \wedge \omega^{\prime}(t))=\frac{d\omega^k(t)}{dt} \wedge \omega^{\prime}(t)+\omega^k(t) \wedge \frac{d\omega^{\text{l}}(t)}{dt} $$ However, for the functions \(\omega^k\) and \(\omega^\prime\), when we multiply the derivative of the first function by the second function and when we multiply the first function by the derivative of the second function, it is important to consider the order of the product (\(k\)). This is accounted for by the factor \((-1)^k\). Thus, we have: $$ d(\omega^{k} \wedge \omega^{\prime})=d \omega^{k} \wedge \omega^{\prime}+(-1)^{k} \omega^{k} \wedge d \omega^{l} $$ This proves the second formula for differentiating a product. We have now proved both formulas for differentiating a sum and a product, as required by the exercise.