Question

find the derivative of the function. y = x^{9/5}+e^{-4x} \frac{dy}{dx}=square

Explanation:

Step1: Apply power - rule for $x^{\frac{9}{5}}$

The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$. For $y_1=x^{\frac{9}{5}}$, we have $y_1^\prime=\frac{9}{5}x^{\frac{9}{5}-1}=\frac{9}{5}x^{\frac{4}{5}}$.

Step2: Apply chain - rule for $e^{-4x}$

The chain - rule states that if $y = e^{u}$ and $u = g(x)$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. For $y_2 = e^{-4x}$, let $u=-4x$, then $\frac{dy_2}{du}=e^{u}$ and $\frac{du}{dx}=-4$. So $\frac{dy_2}{dx}=e^{-4x}\cdot(-4)=-4e^{-4x}$.

Step3: Find the derivative of $y$

Since $y = y_1 + y_2$, by the sum - rule of differentiation $\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$. So $\frac{dy}{dx}=\frac{9}{5}x^{\frac{4}{5}}-4e^{-4x}$.

Answer:

$\frac{9}{5}x^{\frac{4}{5}}-4e^{-4x}$