Question

find $\frac{d}{dp}(a^{2}p^{4}+b^{3}p + c^{5})$. $\frac{d}{dp}(a^{2}p^{4}+b^{3}p + c^{5})=2acdot4p^{3}+3b^{2}cdot p + 5c^{4}$

Explanation:

Step1: Apply sum - rule of differentiation

The derivative of a sum of functions is the sum of their derivatives. So, $\frac{d}{dp}(a^{2}p^{4}+b^{3}p + c^{5})=\frac{d}{dp}(a^{2}p^{4})+\frac{d}{dp}(b^{3}p)+\frac{d}{dp}(c^{5})$.

Step2: Apply power - rule of differentiation

The power - rule states that $\frac{d}{dp}(x^{n})=nx^{n - 1}$. For $\frac{d}{dp}(a^{2}p^{4})$, since $a$ is treated as a constant, $\frac{d}{dp}(a^{2}p^{4})=a^{2}\frac{d}{dp}(p^{4})=a^{2}\times4p^{3}=4a^{2}p^{3}$. For $\frac{d}{dp}(b^{3}p)$, since $b$ is treated as a constant, $\frac{d}{dp}(b^{3}p)=b^{3}\frac{d}{dp}(p)=b^{3}\times1 = b^{3}$. For $\frac{d}{dp}(c^{5})$, since $c$ is a constant, the derivative of a constant is 0. So $\frac{d}{dp}(c^{5}) = 0$.

Step3: Combine the results

$\frac{d}{dp}(a^{2}p^{4}+b^{3}p + c^{5})=4a^{2}p^{3}+b^{3}+0 = 4a^{2}p^{3}+b^{3}$.

Answer:

$4a^{2}p^{3}+b^{3}$