Question

this is similar to section 2.7 problem 30: for the function y = 6x^7+3xe^{2x - 6} find \frac{dy}{dx}. answer: \frac{dy}{dx}=

Explanation:

Step1: Differentiate each term separately

We use the sum - rule of differentiation $\frac{d}{dx}(u + v+w)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}$, where $u = 6x^{7}$, $v=3xe^{2x}$ and $w=-6$.

Step2: Differentiate $u = 6x^{7}$

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $a = 6$ and $n = 7$, we have $\frac{du}{dx}=6\times7x^{7 - 1}=42x^{6}$.

Step3: Differentiate $v = 3xe^{2x}$

Using the product - rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 3x$ and $v = e^{2x}$. First, $\frac{du}{dx}=3$ and $\frac{dv}{dx}=e^{2x}\times2 = 2e^{2x}$. Then $\frac{dv}{dx}=3x\times2e^{2x}+3e^{2x}=3e^{2x}(2x + 1)$.

Step4: Differentiate $w=-6$

Since the derivative of a constant is 0, $\frac{dw}{dx}=0$.

Step5: Combine the derivatives

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=42x^{6}+3e^{2x}(2x + 1)+0$.

Answer:

$42x^{6}+3e^{2x}(2x + 1)$