Answer to Question #257570 in Calculus for shyla

Question

Use an appropriate technique to find the derivative of the following functions:

a. $[ x y^3/ 1 + sec ( y ) ]$ = e^xy

^b. $(√( 3 x ^2 + 4)/ ( x ^2 + 1)^(1/3^))(π^X)$

Answer 1

Solution:

a.

$\dfrac{xy^3}{1+\sec y} = e^{xy}$

On differentiating both sides w.r.t $x$,

$\dfrac{(1+\sec y)(xy^3)'-(xy^3)(1+\sec y)'}{(1+\sec y)^2} = e^{xy}(xy)'$

$\Rightarrow \dfrac{(1+\sec y)(y^3+3xy^2y')-(xy^3)(\sec y \tan y)y'}{(1+\sec y)^2} = e^{xy}(xy'+y)$

$\Rightarrow \dfrac{(y^3+y^3\sec y+3xy^2y'+3xy^2y'\sec y)-(xy^3y'\sec y \tan y)}{(1+\sec y)^2} = xy'e^{xy}+ye^{xy}$

$\Rightarrow y^3+y^3\sec y+y'[3xy^2+3xy^2\sec y-xy^3\sec y \tan y] = xy'e^{xy}(1+\sec y)^2+ye^{xy}(1+\sec y)^2$

$\Rightarrow y'[3xy^2+3xy^2\sec y-xy^3\sec y \tan y] - xy'e^{xy}(1+\sec y)^2=ye^{xy}(1+\sec y)^2-(y^3+y^3\sec y) \\\Rightarrow y'[3xy^2+3xy^2\sec y-xy^3\sec y \tan y - xe^{xy}(1+\sec y)^2]=ye^{xy}(1+\sec y)^2-(y^3+y^3\sec y) \\\Rightarrow y'=\dfrac{ye^{xy}(1+\sec y)^2-(y^3+y^3\sec y)}{3xy^2+3xy^2\sec y-xy^3\sec y \tan y - xe^{xy}(1+\sec y)^2}$

b.

Let $y=\sqrt{\dfrac{3x^2+4}{(x^2+1)^{1/3}}}\times\pi^x$

Taking log to both sides,

$\log y=\log (\sqrt{\dfrac{3x^2+4}{(x^2+1)^{1/3}}}\times\pi^x) \\\Rightarrow \log y=\log (\sqrt{\dfrac{3x^2+4}{(x^2+1)^{1/3}}})+\log(\pi^x) \\\Rightarrow \log y=\dfrac12\log {\dfrac{3x^2+4}{(x^2+1)^{1/3}}}+x\log(\pi) \\\Rightarrow \log y=\dfrac12[\log ({{3x^2+4})-\log{(x^2+1)^{1/3}}}]+x\log(\pi)$

$\\\Rightarrow \log y=\dfrac12[\log ({{3x^2+4})-\dfrac13\log{(x^2+1)}}]+x\log(\pi) \\\Rightarrow \log y=\dfrac12\log ({{3x^2+4})-\dfrac16\log{(x^2+1)}}+x\log(\pi)$

On differentiating both sides w.r.t $x$,

$\\\Rightarrow \dfrac1y.y'=\dfrac12.\dfrac{1}{3x^2+4}.(6x)-\dfrac16.\dfrac1{(x^2+1)}.(2x)+1.\log(\pi) \\\Rightarrow y'=y[\dfrac{3x}{3x^2+4}-\dfrac x{3(x^2+1)}+\log(\pi)] \\\Rightarrow y'=\sqrt{\dfrac{3x^2+4}{(x^2+1)^{1/3}}}\times\pi^x[\dfrac{3x}{3x^2+4}-\dfrac x{3(x^2+1)}+\log(\pi)]$