Calculus Solutions Ex#2.3

 

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Q 01-08
Find $\frac{d y}{d x}$.
 
 
Q.01
$$ y=4 x^{7} $$
Solution:
$$ y=4 x^{7} —-(1)$$ Power rule: $$ \frac{d}{d x}\left[x^{r}\right]=r x^{r-1} \text { where } r \text { in any real number } $$ differentiating (1) w.r.t $x$ $$ \therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left[4 x^{7}\right]=4 \frac{d}{d x}\left[x^{7}\right]=4\left[7 \cdot x^{7-1}\right]=28 x^{6} $$
 
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Q.02
 
$$y=-3 x^{12}$$ differentiating w.r.t $x$ $$ \begin{aligned} \frac{d y}{d x}=\frac{d}{d x}\left[-3 x^{12}\right] & =-3 \frac{d}{d x}\left[x^{12}\right] \\ \\ & =-3\left[12 x^{12-1}\right] \\ \\ & =-36\left[ x^{11}\right] \quad \therefore \frac{d}{d x}\left[x^{r}\right]=r x^{r-1} \end{aligned} $$
 
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Q.03
 
$$ \quad y=3 x^{8}+2 x+1$$ differentiating w.r.t $x$ $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[3 x^{8}+2 x+1\right] & \\ \\ & =\frac{d}{d x}\left[3 x^{8}\right]+\frac{d}{d x}[2 x]+\frac{d}{d x}[1] & \therefore \frac{d}{d x}[c]=0 \\  \\ & =3 \frac{d}{d x}\left[x^{8}\right]+2 \frac{d}{d x}[x]+0  &\text { where } c \text { is any constant } \\ \\ & =3\left[8 x^{7}\right]+2[1] \quad \therefore \frac{d}{d x}[x]=1 &  \\ & =24 x^{7}+2  \end{aligned} $$
 
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Q.04
 
$$\quad y=\frac{1}{2}\left(x^{4}+7\right)$$ differentiating w.r.t $x$ $$ \begin{aligned} \\ \frac{d y}{d x}=\frac{d}{d x}\left[\frac{1}{2}\left(x^{4}+7\right)\right] \\ \\ & =\frac{1}{2} \frac{d}{d x}\left[x^{4}+7\right] & \\ \\ & =\frac{1}{2} [ \frac{d}{d x}\left[x^{4}\right]+ \frac{d}{d x}[7]] & \\ \\ & =\frac{1}{2}\left[4 x^{4-1}+0\right] &  \frac{d[c]}{d x}=0 \\ \\ \frac{d y}{d x} & =2 x^{3} &  c \text { is any constant } \\ & & \frac{d}{d x}\left[x^{r}\right]=r x^{r-1} \end{aligned} $$ ——
 
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Q.05
 
$$y={\pi}^{3}$$ differentiating w.r.t $x$ $$ \frac{d y}{d x}=\frac{d}{d x}\left[{\pi}^{3}\right]=0 $$ since $\pi}^{3}$ is a constant, so its derivative is zero
 
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Q.06
 
$$y=\sqrt{2} x+1 / \sqrt{2}$$ differentiating w.r.t $ x$ $$ \begin{aligned}  \\ \frac{d y}{d x} & =\frac{d}{d x}[\sqrt{2} x+1 / \sqrt{2}]=\frac{d}{d x}[\sqrt{2} x]+\frac{d}{d x}[1 / \sqrt{2}] \\ \\ & =\sqrt{2} \frac{d}{d x}[x]+0 \quad \therefore \frac{d}{d x}[c]=0, c \text { in } \\ \\ \frac{d y}{d x} & =\sqrt{2}(1)=\sqrt{2} \quad \therefore \quad \frac{d}{d x}[x]=1 \end{aligned} $$
 
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Q.07
 
$$ \begin{aligned} y & =-\frac{1}{3}\left(x^{7}+2 x-9\right) \\ y & =-\frac{1}{3} x^{7}-\frac{2}{3} x+3 \end{aligned} $$ differentiating w.r.t $x$ $$ \begin{aligned}  \\ & \frac{d y}{d x}  =\frac{d}{d x}\left[-\frac{1}{3} x^{7}-\frac{2}{3} x+3\right] \\ \\ & =\frac{d}{d x}\left[-\frac{1}{3} x^{7}\right]-\frac{d}{d x}\left[\frac{2}{3} x\right]+\frac{d}{d x}[3] \\ \\ & =-\frac{1}{3} \frac{d}{d x}\left[x^{7}\right]-2 \frac{d}{d x}[x]+0 \quad \therefore \frac{d[c]}{d x}=0 \\ \\ & =-\frac{1}{3}\left(7 x^{7-1}\right)-2(1) \quad \therefore \frac{d}{d x}\left[x^{r}\right]=r x^{r-1} \\ \\ \frac{d y}{d x} & =-7 / 3 x^{6} \end{aligned} $$
 
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Q.08
$$y  =\frac{x^{2}+1}{5}$$
 differentiating w.r.t $ x$
$$ \begin{aligned}  \frac{d y}{d x}  =\frac{d}{d x}\left[\frac{1}{5} x^{2}+1 / 5\right] \\  \frac{d y}{d x}=\frac{d}{d x}\left[1 / 5 x^{2}\right]+\frac{d}{d x}[1 / 5] \\   \frac{d y}{d x}=\frac{1}{5}{d x}\left[x^{2}\right]+0 \quad \therefore \frac{d}{d x}[c]=0 \\  \frac{d y}{d x}=\frac{1}{5}\left(2 x^{2-1}\right) \quad \therefore \frac{d}{d x}\left[x^{r}\right]=r x^{r-1} \\   \end{aligned} $$ $$\frac{d y}{d x}=\frac{2}{5} x$$
 
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Q.09
Find $f^{\prime}(x)$ where $f(x)=x^{-3}+\frac{1}{x^7}$
Solution:

$$f(x)=x^{-3}+\frac{1}{x^7}$$

$$f(x)=x^{-3}+x^{-7}$$

differentiating w.r.t $x$

$$

\begin{aligned}

f^{\prime}(x)=\frac{d}{dx}[x^{-3}+x^{-7}] \\ \\  f^{\prime}(x)=\frac{d}{dx}[x^{-3}]+\frac{d}{dx}[x^{-7}] \\ \\ 

f^{\prime}(x)=-3 x^{-3-1}-7 x^{-7-1} \\ \\ f^{\prime}(x)=-3x^{-4}-7 x^{-8}=-3x^{-4}-7\frac{1}{x^7}

\end{aligned}$$

Q.10

Find $f^{\prime}(x)$ where $f(x)=\sqrt{x}+\frac{1}{x}$

Solution:

$$f(x)=\sqrt{x}+\frac{1}{x}$$

$$f(x)=x^{\frac{1}{2}}+x^{-1}$$

differentiating w.r.t $x$ $$ \begin{aligned} f^{\prime}(x)=\frac{d}{dx}[x^{\frac{1}{2}}+x^{-1}] \\ \\ f^{\prime}(x)=\frac{d}{dx}[x^{\frac{1}{2}}]+\frac{d}{dx}[x^{-1}] \\ \\ f^{\prime}(x)=\frac{1}{2} x^{-1/2}- x^{-2} \\ \\ f^{\prime}(x)=\frac{1}{2x^{1/2}} -\frac{1}{ x^{2}} \end{aligned}$$