Calculus Solutions Ex#2.4

 

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Q.01-04
Compute the derivative of  $f(x)$ by (a) multiplying and then differentiating and (b) using the product rule.
The product rule $$ \frac{d}{d x}[f(x) g(x)]=f(x) \frac{d}{d x}[g(x)]+g(x) \frac{d}{d x}[f(x)] $$ The quotient rule $$ \frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x) \frac{d}{d x}[f(x)]-f(x) \frac{d}{d x}(g(x)]}{[g(x)]^2} $$
Solution:
1.(a )$$ f(x)=(x+1)(2 x-1) $$ $$ f(x)=2 x^2-x+2 x-1=2 x^2+x-1 $$ differentiating w.r.t $x$ $$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left[2 x^2+x-1\right]=2 \frac{d}{d x}\left[x^2\right]+\frac{d}{d x}[x]-\frac{d}{d x}[1] \\ \\& f^{\prime}(x)=2(2)(x)^{2-1}+1-0 \quad \therefore \quad \frac{d}{d x}\left[x^r\right]=rx^{r-1} \\ \\& f^{\prime}(x)=4 x+1 \end{aligned} $$
1(b) $$f(x)=(x+1)(2 x-1)$$
Differentiating w.r.t $x$ (using product rule)
$$ \begin{aligned} f^{\prime}(x) & =(x+1) \frac{d}{d x}(2 x-1)+(2 x-1) \frac{d}{d x}(x+1) \\ \\ & \left.=(x+1)\left[\frac{d}{d x}(2 x)-\frac{d}{d x}(1)\right]+(2 x-1)\left[\frac{d}{d x}[x]+\frac{d}{d x}(1)\right]\right] \\ \\ & =(x+1)[2(1)-0]+(2 x-1)[1+0] \\ \\ & =2 x+2+2 x-1=4 x+1 \end{aligned} $$
2. (a) $$f(x)=\left(3 x^2-1\right)\left(x^2+2\right)$$ $$ \begin{aligned} & f(x)=3 x^4+6 x^2-x^2-2 \\ \\& f(x)=3 x^4+5 x^2-2 \end{aligned} $$ differentiating w.r.t $ x$ $$ \begin{aligned} f^{\prime}(x)=\frac{d}{d x}\left[3 x^4+5 x^2-2\right]=3 \frac{d}{d x}\left[x^4\right]+5 \frac{d}{d x}\left[x^2\right]-\frac{d}{d x}[2] \\  \\f^{\prime}(x)=3(4) x^{4-1}+5(2) x^{2-1}-0 \quad \therefore \frac{d}{d x}\left[x^r\right]=rx^{r-1}  \end{aligned} $$ $$f^{\prime}(x)=12 x^3+10 x$$
(b) $$\quad f(x)=\left(3 x^2-1\right)\left(x^2+2\right)$$ differentiating w.r.t $x$ (using product rule) $$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}\left[3 x^2-1\right]\left[x^2+2\right] \\ \\ & =\left(3 x^2-1\right) \frac{d}{d x}\left[x^2+2\right]+\left[x^2+2\right] \frac{d}{d x}\left[3 x^2-1\right] \\ \\& =\left(3 x^2-1\right)(2 x+0)+\left(x^2+2\right)(6 x-0) \\ \\&=6 x^3-2 x+6 x^3+12 x \\ \\& f^{\prime}(x)=12 x^3+10 x . \end{aligned} $$ 
 
3.(a) $$f(x)=\left(x^2+1\right)\left(x^2-1\right)=x^4-1$$ 
differentiating w.r.t $ x$ 
$$ \begin{aligned}  & f^{\prime}(x)=\frac{d}{d x}\left[x^4-1\right] \\ \\ & =\frac{d}{d x}\left[x^4\right]-\frac{d}{d x}[1] \\ \\ & =4 x^{4-1}-0=4 x^3 \end{aligned} $$
(b)
$$ f(x)=\left(x^2+1\right)\left(x^2-1\right) $$ Taking derivative of $f(x)$ w.r.t $x$ (using product rule). $$ \begin{aligned} & f^{\prime}(x)  =\frac{d}{d x}\left[x^2+1\right]\left[x^2-1\right] \\ \\ & =\left[x^2+1\right] \frac{d}{d x}\left[x^2-1\right]+\left(x^2-1\right) \frac{d}{d x}\left[x^2+1\right] \\ \\ & \left.=\left(x^2+1\right)\left[\frac{d}{d x}\left(x^2\right)-\frac{d}{d x}(1)\right]+\left(x^2-1\right)\left[\frac{d}{d x} x^2\right)+\frac{d}{d x}(1)\right] \\ \\ & =\left(x^2+1\right)(2 x-0)+\left(x^2-1\right)(2 x+0) \\ \\ f^{\prime}(x) & =2 x^3+2 x+2 x^3-2 x=4 x^3 \end{aligned} $$
4. (a) $$ \begin{aligned} & f(x)=\left(x+1\right)\left(x^2-x+1\right) \\ \\ & f(x)=x^3-x^2+x+x^2-x+1 \\ \\& f(x)=x^3+1 \end{aligned} $$ differentiating w.r.t $x$. $$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left[x^3+1\right]=\frac{d}{d x}\left[x^3\right]+\frac{d}{d x}[1] \\ \\& f^{\prime}(x)=3 x^2+0=3 x^2 \quad \therefore \frac{d}{d x}\left[x^r\right]=r x^{r-1} \end{aligned} $$
(b) $$ \begin{aligned} & f(x)=(x+1)\left(x^2-x+1\right) \end{aligned} $$ Now we shall differentiate $f(x)$ w.r.t $x.$ ( using product rule) $$ f^{\prime}(x) =\frac{d}{d x}\left[(x+1)(x^2-x+1)\right] $$ $$ f^{\prime}(x)=(x+1) \frac{d}{d x}\left(x^2-x+1\right)+\left(x^2-x+1\right) \frac{d}{d x}(x+1) $$ $$f^{\prime}(x)=(x+1)(2x-1)+(x^2-x+1) $$ $$f^{\prime}(x)=2x^2+2x-x-1+x^2-x+1 $$ $$f^{\prime}(x)=3x^2$$
Q.5 Find $f^{\prime}(x) $where $$f(x)=\left(3x^2+6\right)\left(2x-\frac{1}{4}\right)$$
Solution: $$f(x)=\left(3x^2+6\right)\left(2x-\frac{1}{4}\right)$$ Now differentiating $f(x)$w.r.t. $x$ (using product rule)  $$ \begin{aligned} f^{\prime} (x)=\frac{d}{d x}\left[(3x^2+6)(2x-\frac{1}{4})\right] \\ \\ f^{\prime} (x)=(3x^2+6)\frac{d}{d x}(2x-\frac{1}{4})+(2x-\frac{1}{4})\frac{d}{dx}(3x^2+6) \\ \\ f^{\prime} (x)=(3x^2+6)(2)+(2x-\frac{1}{4})(6x) \\ \\ f^{\prime} (x)=6x^2+12+12x^2-\frac{3}{2}x \\ \\ f^{\prime} (x)=18x^2-\frac{3}{2}x+12 \end{aligned} $$
Q.6 Find $f^{\prime}(x) $where $$f(x)=\left(2-x-3x^3\right)\left(7+x^5\right)$$
Solution: $$f(x)=\left(2-x-3x^3\right)\left(7+x^5\right)$$
differentiating w.r.t. $x$(using product rule) $$ \begin{aligned}  f^{\prime} (x)=\frac{d}{d x}\left[(2-x-3x^3)(7+x^5)\right] \\ \\ f^{\prime} (x)=(2-x-3x^3)\frac{d}{d x}(7+x^5)+(7+x^5)\frac{d}{dx}(2-x-3x^3) \\ \\f^{\prime} (x)=(2-x-3x^3)(5x^4)+(7+x^5)(-1-9x^2) \\ \\f^{\prime} (x)=10x^4-5x^5-15x^7-7-x^5-63x^2-9x^7 \\ \\f^{\prime} (x)=-24x^7-6x^5+10x^4-63x^2-7 \end{aligned} $$