Question

$\frac{d}{dx}(x^{6}+3x^{4}+7x^{2})=$

Explanation:

Step1: Apply power - rule of differentiation

The power - rule states that $\frac{d}{dx}(x^n)=nx^{n - 1}$ and for a sum of functions $y = f(x)+g(x)+h(x)$, $\frac{d}{dx}(y)=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))+\frac{d}{dx}(h(x))$.
For $y=x^6 + 3x^4+7x^2$, we have $\frac{d}{dx}(x^6 + 3x^4+7x^2)=\frac{d}{dx}(x^6)+3\frac{d}{dx}(x^4)+7\frac{d}{dx}(x^2)$.

Step2: Differentiate each term

$\frac{d}{dx}(x^6)=6x^{6 - 1}=6x^5$, $\frac{d}{dx}(x^4)=4x^{4 - 1}=4x^3$, $\frac{d}{dx}(x^2)=2x^{2 - 1}=2x$.
So, $3\frac{d}{dx}(x^4)=3\times4x^3 = 12x^3$ and $7\frac{d}{dx}(x^2)=7\times2x=14x$.

Step3: Combine the results

$\frac{d}{dx}(x^6 + 3x^4+7x^2)=6x^5+12x^3 + 14x$.

Answer:

$6x^5+12x^3 + 14x$