1. Find \( \frac{d y}{d x} \) for each of the following functions. \( \begin{array}{lll}\text { (a) } y=4 x^{2}(5 x+3) & \text { (b) } y=-2 x^{3}(x+1) & \text { (c) } y=x^{2}(1-4 x)^{4} \\ \text { (d) } y=x^{2} \sqrt{1-2 x^{2}} & \text { (e) } y=(4 x-3)(2 x+7)^{6} & \text { (f) } y=(x+5)^{3}(x-4)^{4}\end{array} \)

**(a)** Given \[ y=4x^2(5x+3), \] let \[ u=4x^2,\quad v=5x+3. \] Then, \[ u'=8x,\quad v'=5. \] By the product rule, \[ \frac{dy}{dx}=u'v+uv'=8x(5x+3)+4x^2\cdot 5=40x^2+24x+20x^2. \] Simplify: \[ \frac{dy}{dx}=60x^2+24x. \] --- **(b)** For \[ y=-2x^3(x+1), \] set \[ u=-2x^3,\quad v=x+1. \] Then, \[ u'=-6x^2,\quad v'=1. \] Thus, \[ \frac{dy}{dx}=u'v+uv'=-6x^2(x+1)-2x^3. \] Expanding, \[ \frac{dy}{dx}=-6x^3-6x^2-2x^3=-8x^3-6x^2. \] --- **(c)** For \[ y=x^2(1-4x)^4, \] choose \[ u=x^2,\quad v=(1-4x)^4. \] Then, \[ u'=2x. \] Using the chain rule for \( v \), let \( w=1-4x \) so that \[ v=w^4,\quad \frac{dv}{dw}=4w^3,\quad \frac{dw}{dx}=-4. \] Thus, \[ v'=\frac{dv}{dx}=4(1-4x)^3\cdot (-4)=-16(1-4x)^3. \] Now, \[ \frac{dy}{dx}=2x(1-4x)^4-16x^2(1-4x)^3. \] Factor out the common terms \( 2x(1-4x)^3 \): \[ \frac{dy}{dx}=2x(1-4x)^3\Big[(1-4x)-8x\Big]=2x(1-4x)^3(1-12x). \] --- **(d)** For \[ y=x^2\sqrt{1-2x^2}, \] rewrite the square root as a power: \[ y=x^2(1-2x^2)^{\frac{1}{2}}. \] Let \[ u=x^2,\quad v=(1-2x^2)^{\frac{1}{2}}, \] so that \[ u'=2x. \] For \( v \), using the chain rule with \( w=1-2x^2 \): \[ v=w^{\frac{1}{2}},\quad \frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}},\quad \frac{dw}{dx}=-4x. \] Thus, \[ v'=\frac{1}{2}(1-2x^2)^{-\frac{1}{2}}(-4x)=-2x(1-2x^2)^{-\frac{1}{2}}. \] Apply the product rule: \[ \frac{dy}{dx}=2x(1-2x^2)^{\frac{1}{2}}-2x^3(1-2x^2)^{-\frac{1}{2}}. \] Factor out \( 2x(1-2x^2)^{-\frac{1}{2}} \): \[ \frac{dy}{dx}=2x(1-2x^2)^{-\frac{1}{2}}\Big[(1-2x^2)-x^2\Big]=2x(1-2x^2)^{-\frac{1}{2}}(1-3x^2). \] Alternatively, \[ \frac{dy}{dx}=\frac{2x(1-3x^2)}{\sqrt{1-2x^2}}. \] --- **(e)** For \[ y=(4x-3)(2x+7)^6, \] take \[ u=4x-3,\quad v=(2x+7)^6. \] Then, \[ u'=4, \] and for \( v \) (using the chain rule with \( w=2x+7 \)): \[ v= w^6,\quad \frac{dv}{dw}=6w^5,\quad \frac{dw}{dx}=2, \] so, \[ v'=6(2x+7)^5\cdot 2=12(2x+7)^5. \] By the product rule, \[ \frac{dy}{dx}=4(2x+7)^6+12(4x-3)(2x+7)^5. \] Factor out \( (2x+7)^5 \): \[ \frac{dy}{dx}=(2x+7)^5\Big[4(2x+7)+12(4x-3)\Big]. \] Simplify the bracket: \[ 4(2x+7)=8x+28,\quad 12(4x-3)=48x-36, \] so, \[ 8x+28+48x-36=56x-8=8(7x-1). \] Thus, \[ \frac{dy}{dx}=8(2x+7)^5(7x-1). \] --- **(f)** For \[ y=(x+5)^3(x-4)^4, \] let \[ u=(x+5)^3,\quad v=(x-4)^4. \] Then, \[ u'=3(x+5)^2,\quad v'=4(x-4)^3. \] Apply the product rule: \[ \frac{dy}{dx}=3(x+5)^2(x-4)^4+4(x+5)^3(x-4)^3. \] Factor out the common factor \( (x+5)^2(x-4)^3 \): \[ \frac{dy}{dx}=(x+5)^2(x-4)^3\Big[3(x-4)+4(x+5)\Big]. \] Simplify the bracket: \[ 3(x-4)=3x-12,\quad 4(x+5)=4x+20, \] so, \[ 3x-12+4x+20=7x+8. \] Thus, \[ \frac{dy}{dx}=(x+5)^2(x-4)^3(7x+8). \]