4.5 Questions related to derivation of common functions

Question 1. Differentiate $y = \cos(ln(4x^3-5x+3))$.

Solution: Using chain rule

$$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\cos (\ln (4{x^3} - 5x + 3))} \right)$$

We know that, $$\frac{d}{dx} \cos x = -sin x$$$$\frac{{dy}}{{dx}} = - \sin (\ln (4{x^3} - 5x + 3))\frac{d}{{dx}}\ln (4{x^3} - 5x + 3)$$

We know that, $$\frac{d}{dx}{\log _e}x = \frac{1}{x \ln e}= \frac{1}{x}$$$$\frac{{dy}}{{dx}} = - \frac{{\sin (\ln (4{x^3} - 5x + 3))}}{{4{x^3} - 5x + 3}}\frac{d}{{dx}}(4{x^3} - 5x + 3)$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{{dx}}k = 0\quad where\,k = constant$$

Using sum and difference and scalar rule,

$$\frac{{dy}}{{dx}} = - \frac{{\sin (\ln (4{x^3} - 5x + 3))}}{{4{x^3} - 5x + 3}}\left( {4\frac{d}{{dx}}{x^3} - 5\frac{d}{{dx}}x + \frac{d}{{dx}}3)} \right)$$$$\frac{{dy}}{{dx}} = - \frac{{\sin (\ln (4{x^3} - 5x + 3))}}{{4{x^3} - 5x + 3}}\left( {4(3){x^{3 - 1}} - 5{x^{1 - 1}} + 0} \right)$$$$\frac{{dy}}{{dx}} = - \frac{{\sin (\ln (4{x^3} - 5x + 3))}}{{4{x^3} - 5x + 3}}\left( {12{x^2} - 5} \right)$$

Question 2. Differentiate $y=10x^5+9x^4-8x^3+7x^2-6$.

Solution: Using sum and difference and scalar rule,

$$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {10{x^5} + 9{x^4} - 8{x^3} + 7{x^2} - 6} \right)$$$$\frac{dy}{dx}=10\frac{d}{dx}x^5+9\frac{d}{dx}x^4-8\frac{d}{dx}x^3+7\frac{d}{dx}x^2-\frac{d}{dx}6$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{{dx}}k = 0\quad where\,k = constant$$$$\frac{dy}{dx}=10(5)x^{5-1}+9(4)x^{4-1}-8(3)x^{3-1}+7(2)x^{2-1}-0$$$$\frac{dy}{dx}=50x^4+36x^3-24x^2+14x$$

Question 3. Differentiate $y=4e^x (1+ln\,x) tan\,x$.

Solution: Using product rule and scalar rule,

$$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {4{e^x}(1 + ln{\kern 1pt} x)\tan x} \right)$$$$\frac{{dy}}{{dx}} = \left( {\frac{d}{{dx}}4{e^x}} \right)\left( {(1 + ln{\kern 1pt} x)\tan x} \right) + 4{e^x}\left( {\frac{d}{{dx}}(1 + ln{\kern 1pt} x)} \right){\kern 1pt} \tan x + 4{e^x}(1 + ln{\kern 1pt} x)\frac{d}{{dx}}\tan x$$

We know that,

$$\frac{d}{{dx}}{e^x} = {e^x}$$$$\frac{d}{dx} \tan x = \sec ^2 x$$

Using sum rule,

$$\frac{{dy}}{{dx}} = 4{e^x}(1 + ln{\kern 1pt} x)\tan x + 4{e^x}\left( {\frac{d}{{dx}}(1) + \frac{d}{{dx}}ln{\kern 1pt} x} \right){\kern 1pt} \tan x + 4{e^x}(1 + ln{\kern 1pt} x){\sec ^2}x$$

We know that,

$$\frac{d}{{dx}}k = 0\quad where\,k = constant$$$$\frac{d}{dx}{\log _e}x = \frac{1}{x ln e}= \frac{1}{x}$$$$\frac{{dy}}{{dx}} = 4{e^x}(1 + ln{\kern 1pt} x)\tan x + 4{e^x}\left( {0 + \frac{1}{x}} \right){\kern 1pt} \tan x + 4{e^x}(1 + ln{\kern 1pt} x){\sec ^2}x$$$$\frac{{dy}}{{dx}} = 4{e^x}(1 + ln{\kern 1pt} x)\tan x + 4{e^x}\left( {\frac{1}{x}} \right){\kern 1pt} \tan x + 4{e^x}(1 + ln{\kern 1pt} x){\sec ^2}x$$$$\frac{{dy}}{{dx}} = 4{e^x}\left( {(1 + ln{\kern 1pt} x)\tan x + \frac{{\tan x}}{x}{\kern 1pt} + (1 + ln{\kern 1pt} x){{\sec }^2}x} \right)$$

Question 4. Differentiate $y=\frac{cot^{-1}x}{3^x}$.

Solution: Using quotient rule,

$$\frac{dy}{dx}= \frac{{{3^x}\frac{d}{{dx}}{{\cot }^{ - 1}}x - {{\cot }^{ - 1}}x\frac{d}{{dx}}{3^x}}}{{{{\left( {{3^x}} \right)}^2}}}$$

We know that,

$$\frac{d}{dx} cot^{-1} x = -\frac {1}{1+x^2}$$$$\frac{d}{dx} a^x = a^x log a$$$$\frac{{dy}}{{dx}} = \frac{{{3^x}\left( { - \frac{1}{{1 + {x^2}}}} \right) - {{\cot }^{ - 1}}x\left( {{3^x}\log 3} \right)}}{{{3^{2x}}}}$$$$\frac{{dy}}{{dx}} = \frac{{ - {3^x} - \left( {{3^x}\log 3} \right){{\cot }^{ - 1}}x\left( {1 + {x^2}} \right)}}{{{3^{2x}}\left( {1 + {x^2}} \right)}}$$$$\frac{{dy}}{{dx}} = - \frac{1}{{{3^x}\left( {1 + {x^2}} \right)}} - \frac{{\left( {\log 3} \right){{\cot }^{ - 1}}x}}{{{3^x}}}$$

Question 5. Let $u=x^3$ and $y=\cosh u$ then find $\frac{{dy}}{{dx}}$

Solution: $$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\cosh u} \right)$$

Using chain rule,

$$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}$$

We know that $$\frac{d}{dx} \cosh x = sinh x$$$$\frac{{dy}}{{dx}} = \sinh \,u\;\frac{{du}}{{dx}}\quad \ldots (A)$$

Given: $$u=x^3$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$ $$\frac{{du}}{{dx}} = 3{x^{3 - 1}} = 3{x^2}$$

Substitute value of $u = {x^3}$ and $\frac{{du}}{{dx}} = 3{x^2}$ in $(A)$ we get,

$$\frac{{dy}}{{dx}} = \sinh \,{x^3}(3{x^2}) = 3{x^2}\sinh \,{x^3}$$

Question 6. Find the deriative of $y =2 tanh x + 4 sec x$

Solution: $$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {2\tanh x + 4\sec x} \right)$$

Using sum and scalar rule,

$$\frac{{dy}}{{dx}} = 2\frac{d}{{dx}}\tanh x + 4\frac{d}{{dx}}\sec x$$

We know that, $$\frac{d}{{dx}}\tanh x = {\operatorname{sech} ^2}x$$$$\frac{d}{dx} \sec x = \sec x \; \tan x$$$$\frac{{dy}}{{dx}} = 2\sec {h^2}x + 4\sec x\,{\kern 1pt} \tan x$$

Question 7. Find the derivative of $y =u^2 \cos ech u$ where $u=4x^2$

Solution: $$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{u^2}\ cos ech {\kern 1pt} \,u} \right)$$

Using product rule,

$$\frac{{dy}}{{dx}} = \ cos ech {\kern 1pt} \,u\frac{d}{{dx}}{u^2} + {u^2}\frac{d}{{dx}}\ cos ech {\kern 1pt} \,u$$

We know that, $$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{dx} \cos ech x = -\cos ech x \;coth x$$

Using chain rule,

$$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}$$$$\frac{{dy}}{{dx}} = \cos ech\left( u \right)\left( {2{u^{2 - 1}}} \right)\frac{{du}}{{dx}} + {u^2}\left[ { - \cos ech\left( u \right)\coth \left( u \right)} \right]\frac{{du}}{{dx}}$$$$\frac{{dy}}{{dx}} = 2u\ cos ech{\kern 1pt} \,u\frac{{du}}{{dx}} - {u^2}\left( {\ cos ech\,u\,\coth \,u} \right)\frac{{du}}{{dx}}\; \ldots (A)$$

Given,$$ u = 4{x^2} $$$$ \frac{{du}}{{dx}} = 4(2){x^{2 - 1}} = 8x $$Substitute $u = 4{x^2}$ and $\frac{{du}}{{dx}} = 8x$ in $(A)$,

We get,$$\frac{{dy}}{{dx}} = 2(4{x^2})\left( {\cos ech{\kern 1pt} \,(4{x^2})} \right)(8x) - {(4{x^2})^2}\left( {\cos ech\,(4{x^2})\,\coth \,(4{x^2})} \right)(8x)$$$$\frac{{dy}}{{dx}} = 64{x^3}\left( {\cos ech{\kern 1pt} \,(4{x^2})} \right) - 128{x^5}\left( {\cos ech\,(4{x^2})\,\coth \,(4{x^2})} \right)$$