differentiate the following with regard to $x$:\n3.2.1 $y = x\\cos5x$\n3.2.2 $y = \\sqrt{4x - x^{2}}$\n3.2.3…

Explanation:

Step1: Apply product - rule for 3.2.1

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $y^\prime=u^\prime v + uv^\prime$. Here, $u = x$ and $v=\cos5x$. We know that $u^\prime = 1$ and $v^\prime=-5\sin5x$. $y^\prime=1\times\cos5x+x\times(- 5\sin5x)=\cos5x - 5x\sin5x$

Step2: Apply chain - rule for 3.2.2

First, rewrite $y=(4x - x^{2})^{\frac{1}{2}}$. Let $u = 4x - x^{2}$, so $y = u^{\frac{1}{2}}$. Then $\frac{du}{dx}=4 - 2x$ and $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$. $\frac{dy}{dx}=\frac{1}{2}(4x - x^{2})^{-\frac{1}{2}}\times(4 - 2x)=\frac{2 - x}{\sqrt{4x - x^{2}}}$

Step3: Apply chain - rule for 3.2.3

Let $u=\sin e^{3x}$, so $y = \ln u$. Then $\frac{dy}{du}=\frac{1}{u}$ and for $u=\sin e^{3x}$, let $t = e^{3x}$, so $u=\sin t$. $\frac{du}{dt}=\cos t$ and $\frac{dt}{dx}=3e^{3x}$. By the chain - rule, $\frac{du}{dx}=\frac{du}{dt}\times\frac{dt}{dx}=\cos e^{3x}\times3e^{3x}$. And $\frac{dy}{dx}=\frac{1}{\sin e^{3x}}\times\cos e^{3x}\times3e^{3x}=3e^{3x}\cot e^{3x}$

Answer:

3.2.1: $\cos5x - 5x\sin5x$ 3.2.2: $\frac{2 - x}{\sqrt{4x - x^{2}}}$ 3.2.3: $3e^{3x}\cot e^{3x}$