Given y = \frac{x^4e^{3x}(2x^2 + 3x + 1)}{5^{\ln x}(2+x)^4} Find \frac{dy}{dx}

Question:

Given {eq}y = \frac{x^4e^{3x}(2x^2 + 3x + 1)}{5^{\ln x}(2+x)^4} {/eq}

Find {eq}\frac{dy}{dx} {/eq}

Derivative by Quotient Rule:

The quotient rule is a rule for derivation and it is one function is divided by other function. It follows the definition of the limit of differentiation and rule is given by {eq}\displaystyle \frac{d}{du} \frac{f(u)}{g(u)} = \frac{g(u)f'(u)-f(u)g'(u)}{(g(u))^2}. {/eq}

Here we apply following important notations to solve the derivative:

1. The product rule: {eq}\displaystyle \left(f\cdot g\right)'=f'\cdot g+f\cdot g'. {/eq}

2. The power rule: {eq}\displaystyle \frac{d}{du}\left(u^a\right)=a\cdot u^{a-1}. {/eq}

3. The chain rule: {eq}\displaystyle \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}. {/eq}

4. The sum/difference rule: {eq}\displaystyle \left(f\pm g\right)'=f'\pm g'. {/eq}

5. Eject the constant out: {eq}\displaystyle \left(a\cdot f\right)'=a\cdot f'. {/eq}

6. Common derivative: {eq}\displaystyle \frac{d}{du}\left(e^u\right)=e^u. {/eq}

7. Common derivative: {eq}\displaystyle \frac{d}{du}\left(u\right)=1. {/eq}

8. Derivative of a constant: {eq}\displaystyle \frac{d}{du}\left(a\right)=0. {/eq}

9. The exponent rule: {eq}\displaystyle a^b=e^{b\ln \left(a\right)}. {/eq}

Solving for $$\displaystyle y = \frac{x^4e^{3x}(2x^2 + 3x + 1)}{5^{\ln x}(2+x)^4}$$

Take derivative both sides.

$$\displaystyle \frac{d}{dx}y = \frac{d}{dx}\left(\frac{x^4e^{3x}\left(2x^2+3x+1\right)}{5^{\ln \left(x\right)}\left(2+x\right)^4}\right)$$

Apply the quotient rule.

\begin{align*} \displaystyle y' &= \frac{\frac{d}{dx}\left(x^4e^{3x}\left(2x^2+3x+1\right)\right)5^{\ln \left(x\right)}\left(2+x\right)^4-\frac{d}{dx}\left(5^{\ln \left(x\right)}\left(2+x\right)^4\right)x^4e^{3x}\left(2x^2+3x+1\right)}{\left(5^{\ln \left(x\right)}\left(2+x\right)^4\right)^2}\\ \displaystyle y' &= \frac{y_1 5^{\ln \left(x\right)}\left(2+x\right)^4-y_2 x^4e^{3x}\left(2x^2+3x+1\right)}{\left(5^{\ln \left(x\right)}\left(2+x\right)^4\right)^2}\\ \end{align*}

Now, we are going to solving for {eq}\displaystyle y_1 = \frac{d}{dx}\left(x^4e^{3x}\left(2x^2+3x+1\right)\right) {/eq}

Apply the product rule.

{eq}\displaystyle y_1 = \frac{d}{dx}\left(x^4\right)e^{3x}\left(2x^2+3x+1\right)+\frac{d}{dx}\left(e^{3x}\left(2x^2+3x+1\right)\right)x^4 {/eq}

Apply the product rule.

{eq}\displaystyle y_1 = \frac{d}{dx}\left(x^4\right)e^{3x}\left(2x^2+3x+1\right)+\left(\frac{d}{dx}\left(e^{3x}\right)\left(2x^2+3x+1\right)+\frac{d}{dx}\left(2x^2+3x+1\right)e^{3x}\right)x^4 {/eq}

Apply the sum/difference rule.

{eq}\displaystyle y_1 = \frac{d}{dx}\left(x^4\right)e^{3x}\left(2x^2+3x+1\right)+\left(\frac{d}{dx}\left(e^{3x}\right)\left(2x^2+3x+1\right)+\left(\frac{d}{dx}\left(2x^2\right)+\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(1\right)\right)e^{3x}\right)x^4 {/eq}

Eject the constant out.

{eq}\displaystyle y_1 = \frac{d}{dx}\left(x^4\right)e^{3x}\left(2x^2+3x+1\right)+\left(\frac{d}{dx}\left(e^{3x}\right)\left(2x^2+3x+1\right)+\left(2\frac{d}{dx}\left(x^2\right)+3\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)\right)e^{3x}\right)x^4 {/eq}

Apply the power rule and common derivative.

{eq}\displaystyle y_1 = 4x^3e^{3x}\left(2x^2+3x+1\right)+\left(6e^{3x}x^2+13e^{3x}x+6e^{3x}\right)x^4 {/eq}

Simplify:

{eq}\displaystyle y_1 = 6e^{3x}x^6+21e^{3x}x^5+18e^{3x}x^4+4e^{3x}x^3 {/eq}

Now, we are going to solving for {eq}\displaystyle y_2 = \frac{d}{dx}\left(5^{\ln \left(x\right)}\left(2+x\right)^4\right) {/eq}

Apply the product rule.

{eq}\displaystyle y_2 = \frac{d}{dx}\left(5^{\ln \left(x\right)}\right)\left(2+x\right)^4+\frac{d}{dx}\left(\left(2+x\right)^4\right)5^{\ln \left(x\right)} {/eq}

Apply the exponent rule.

{eq}\displaystyle y_2 = \frac{d}{dx}\left(e^{\ln \left(x\right)\ln \left(5\right)}\right)\left(2+x\right)^4+\frac{d}{dx}\left(\left(2+x\right)^4\right)5^{\ln \left(x\right)} {/eq}

Apply the chain rule.

{eq}\displaystyle y_2 = \frac{d}{du}\left(e^u\right)\frac{d}{dx}\left(\ln \left(x\right)\ln \left(5\right)\right)\left(2+x\right)^4+\frac{d}{dv}\left(v^4\right)\frac{d}{dx}\left(2+x\right)5^{\ln \left(x\right)} {/eq}

Eject the constant out.

{eq}\displaystyle y_2 = \frac{d}{du}\left(e^u\right)\ln \left(5\right)\frac{d}{dx}\left(\ln \left(x\right)\right)\left(2+x\right)^4+\frac{d}{dv}\left(v^4\right)\frac{d}{dx}\left(2+x\right)5^{\ln \left(x\right)} {/eq}

Apply the sum/difference rule.

{eq}\displaystyle y_2 = \frac{d}{du}\left(e^u\right)\ln \left(5\right)\frac{d}{dx}\left(\ln \left(x\right)\right)\left(2+x\right)^4+\frac{d}{dv}\left(v^4\right)\left(\frac{d}{dx}\left(2\right)+\frac{d}{dx}\left(x\right)\right)5^{\ln \left(x\right)} {/eq}

Apply the common derivative.

{eq}\displaystyle y_2 =\left(e^u\frac{\ln \left(5\right)}{x}\right)\left(2+x\right)^4+\left(4v^3\cdot 1\right)5^{\ln \left(x\right)} {/eq}

Put in the value of u and v.

{eq}\displaystyle y_2 =\ln \left(5\right)x^{\ln \left(5\right)-1}\left(2+x\right)^4+4\left(2+x\right)^3\cdot 5^{\ln \left(x\right)} {/eq}

Put in the value of {eq}y_1 {/eq} and {eq}y_2. {/eq}

\begin{align*} \displaystyle y' &= \frac{\left(6e^{3x}x^6+21e^{3x}x^5+18e^{3x}x^4+4e^{3x}x^3\right)5^{\ln \left(x\right)}\left(2+x\right)^4-\left(\ln \left(5\right)x^{\ln \left(5\right)-1}\left(2+x\right)^4+4\left(2+x\right)^3\cdot 5^{\ln \left(x\right)}\right)x^4e^{3x}\left(2x^2+3x+1\right)}{\left(5^{\ln \left(x\right)}\left(2+x\right)^4\right)^2}\\ \displaystyle y' &= \frac{5^{\ln \left(x\right)}\left(6e^{3x}x^6+21e^{3x}x^5+18e^{3x}x^4+4e^{3x}x^3\right)\left(2+x\right)^4-e^{3x}x^4\left(\ln \left(5\right)x^{\ln \left(5\right)-1}\left(2+x\right)^4+4\left(2+x\right)^3\cdot 5^{\ln \left(x\right)}\right)\left(2x^2+3x+1\right)}{\left(5^{\ln \left(x\right)}\left(2+x\right)^4\right)^2}. \end{align*}