find \( \frac{d y}{d x} \) 3) \( y=\frac{\left(3 x^{2}-5 x\right)^{10}}{\sqrt{x}} \)

Let \[ y=\frac{\left(3x^2-5x\right)^{10}}{\sqrt{x}}. \] We first rewrite the function as \[ y=\left(3x^2-5x\right)^{10}\,x^{-1/2}. \] Define \[ u(x)=\left(3x^2-5x\right)^{10} \quad \text{and} \quad v(x)=x^{-1/2}. \] We will use the product rule: \[ \frac{dy}{dx}=u'(x)v(x)+u(x)v'(x). \] **Step 1.** Differentiate \( u(x) \). By the chain rule, \[ u'(x)=10\left(3x^2-5x\right)^9\cdot\frac{d}{dx}\left(3x^2-5x\right). \] Differentiate the inner function: \[ \frac{d}{dx}\left(3x^2-5x\right)=6x-5. \] Thus, \[ u'(x)=10\left(3x^2-5x\right)^9(6x-5). \] **Step 2.** Differentiate \( v(x)=x^{-1/2} \). We have: \[ v'(x)=-\frac{1}{2}x^{-3/2}. \] **Step 3.** Apply the product rule. Substitute \( u'(x) \) and \( v'(x) \) into the product rule: \[ \frac{dy}{dx}=10\left(3x^2-5x\right)^9(6x-5)\,x^{-1/2}-\frac{1}{2}\left(3x^2-5x\right)^{10}\,x^{-3/2}. \] **Step 4.** Factor common terms. Notice that both terms share a factor of \(\left(3x^2-5x\right)^9\) and a power of \( x^{-3/2} \). Write \( x^{-1/2} \) as \( x^{-3/2}\cdot x \). Then, \[ \frac{dy}{dx}=\left(3x^2-5x\right)^9\,x^{-3/2}\left[10(6x-5)x-\frac{1}{2}\left(3x^2-5x\right)\right]. \] Simplify the expression inside the brackets. First, compute: \[ 10x(6x-5)=60x^2-50x. \] Thus, the bracket becomes: \[ 60x^2-50x-\frac{1}{2}\left(3x^2-5x\right)=60x^2-50x-\frac{3}{2}x^2+\frac{5}{2}x. \] Combine like terms: \[ 60x^2-\frac{3}{2}x^2=\frac{120-3}{2}x^2=\frac{117}{2}x^2, \] \[ -50x+\frac{5}{2}x=\frac{-100+5}{2}x=\frac{-95}{2}x. \] Thus, the expression inside the brackets is: \[ \frac{117}{2}x^2-\frac{95}{2}x=\frac{x\left(117x-95\right)}{2}. \] **Step 5.** Write the final answer. Substitute back: \[ \frac{dy}{dx}=\left(3x^2-5x\right)^9\,x^{-3/2}\cdot\frac{x\left(117x-95\right)}{2}. \] Simplify \( x^{-3/2}\cdot x=x^{-1/2} \): \[ \frac{dy}{dx}=\frac{\left(3x^2-5x\right)^9\left(117x-95\right)}{2x^{1/2}}, \] or equivalently, \[ \frac{dy}{dx}=\frac{\left(3x^2-5x\right)^9\left(117x-95\right)}{2\sqrt{x}}. \]