# 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}

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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 =... 