Question

Find the derivatives of the following functions with respect to x.

𝑥³ + 𝑦³ = 3                                                                                                                   

𝑦 = (sin 𝑥)^{𝑡𝑎𝑛𝑥}

Answer 1

ANSWER

a) If ${ x }^{ 3 }+{ y }^{ 3 }=3$ , then $y(x)={ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }=f(g(x)) }.\ g(x)=3-{ x }^{ 3 }\ ,\ f(x)={ x }^{ 1/3 }$ . By the Chain Rule, we have $y'(x)=f'(g(x))\cdot g'(x)$ . $g'(x)=-3{ x }^{ 2 }\quad ,\ f'(x)=\frac { 1 }{ 3 } { x }^{ -\frac { 2 }{ 3 } }$ . Hence

$y'(x)=-\frac { 1 }{ 3 } { \left( 3-{ x }^{ 3 } \right) }^{ -\frac { 2 }{ 3 } }\cdot 3{ x }^{ 2 }=-{ \left( 3-{ x }^{ 3 } \right) }^{ -\frac { 2 }{ 3 } }\cdot \ { x }^{ 2 }$ .

b) Since ${ a }^{ b }={ e }^{ b\ lna }\quad (a>0)$ , then $y(x)={ { e }^{ (\tan { x } )\cdot \ln { (\sin { x) } } \quad } }$.So, by the Chain Rule:

$y'(x)={ e }^{ (\tan { x } )\cdot \ln { (\sin { x) } } \ }\left[ (\tan { x } )\cdot \ln { (\sin { x) } } \right] '.$

$\left[ (\tan { x } )\cdot \ln { (\sin { x) } } \right] '=$

$={ (\sec { x) } }^{ 2 }\cdot \ln { (\sin { x) } } +(\tan { x } )\cdot \left[ \ \frac { \cos { x } }{ \sin { x } } \right] =$

$= { (\sec { x) } }^{ 2 }\cdot \ln { (\sin { x) } } +1.$ Therefore,

$\quad y'(x)=({ \sin { x) } }^{ \tan { x } }\cdot \left[ { (\sec { x) } }^{ 2 }\cdot \ln { (\sin { x) } } +1 \right]$