Calculus Solutions Ex#2.6

  [embeddoc url=”https://genesismath.com/wp-content/uploads/2023/04/EXERCISE-2.6.pdf” viewer=”google”] Q.01 Solution: Given that $f^{\prime}(0)=2, g(0)=0$ and $g^{\prime}(0)=3$  find  $(f \circ g)^{\prime}(2).$  $$ \begin{aligned}  \text {Formula} (f \circ g)^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x) \quad \text { Put } x=0, \end{aligned}$$  $$ \begin{aligned}(f \circ g)^{\prime}(0) & =f^{\prime}(g(0)) g^{\prime}(0) & & \\ \\ & =f^{\prime}(0) \cdot(3) & \therefore g(0) =0 \\ \\ & =2(3)=6 & & … Read more

Calculus Solutions Ex#2.5

  Fomulas: Derivatives of trigonometric functions $$ \begin{array}{ll} \frac{d}{d x}[\sin x]=\cos x & \frac{d}{d x}[\cos x]=-\sin x \\ \\ \frac{d}{d x}[\tan x]=\sec ^2 x & \frac{d}{d x}[\sec x]=\sec x \tan x \\ \\ \frac{d}{d x}[\cot x]=-\csc x^2 & \frac{d}{d x}[\csc x]=-\csc x \cot x \end{array} $$ Q.01 Find $f^{\prime}(x)$ where $f(x)=4 \cos x+2 \sin x$ … Read more

Calculus Solutions Ex#2.4

  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 … Read more

Calculus Solutions Ex#2.3

  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} … Read more

Calculus Solutions Ex#2.2

  Q.07 Solution: Given $$ \begin{aligned} & m=f^{\prime}(3)=5 \\ \\ & x_{1}=3, \quad y_{1}=f(3)=-1 \end{aligned} $$ Equation of tangent line $$ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ \\ y-(-1) & =5(x-3) \\ \\ y+1 & =5(x-3) \\ \\ y & +1=5 x-15 \\ \\ y & =5 x-15-1 \\ \\ y & =5 x-16 \end{aligned} $$ … Read more

Calculus Solutions EX#2.1

  Q.01 Solution: $$\quad t=10s$$ We shall find the instantaneous velocity  when $t=10s$From the graph, we see thatat $t_1=5 s, \quad s_1=10 m$at $t_2=15 s, \quad s_2=50 m$ Instantaneous velocity at $t=10s$ is equal to the slope of the tangent line at $t=10.$$$m_{tan}=\frac{s_2-s_1}{t_2-t_1}$$ $$m_{tan}=\frac{50-10}{15-5}$$ $$m_{tan}=\frac{40}{10}=4 m/s$$ Q.02 Solution: We shall find instantaneous velocity at $t=4 … Read more