Answer to Question #254289 in Differential Equations for Nazmul

Question

1. Differentiate of the following functions with respect to x:

i) 𝑆𝑖𝑛𝑥∙ ln(5𝑥)

ii) √𝑐𝑜𝑠√𝑥

iii) 𝑒 ln (𝑡𝑎𝑛5𝑥)

iv) 𝑆𝑖𝑛2 {ln(𝑠𝑒𝑐𝑥)}

v) ln (𝑡𝑎𝑛𝑥) �

Answer 1

$(\sin(x))'=\cos(x)$

$(\cos(x))'=-\sin(x)$

$(\ln(x))'=\frac{1}{x}$

$(\sqrt{x})'=\frac{1}{2}\cdot\frac{1}{\sqrt{x}}$

$(e^{x})'=e^{x}$

$(\tan(x))'=\frac{1}{\cos^2(x)}$

$(C\cdot x)'=C$

Let, $h(x)=f(g(x)), h'(x)=f'(x)\cdot g'(x)$ (*)

Use(*),

$(\sec(x))'=(\frac{1}{\cos(x)})'=(-\frac{1}{\cos^{2}(x)})\cdot(-\sin(x))=\frac{\sin(x)}{\cos^2(x)}$ ,

where $f(x)=\frac{1}{\cos(x)}, g(x)=\cos(x)$ ).

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1) $(\sin(x)\cdot\ln(5x))'.$

$(f\cdot g)'=f'g+fg'$ (1)

Use (1),

$(\sin(x)\cdot\ln(5x))'=(\sin(x))'\cdot\ln(5x)+\sin(x)\cdot(\ln(5x))'=$

$=\cos(x)\cdot\ln(5x)+\sin(x)\cdot\frac{1}{5x}\cdot5=\cos(x)\cdot\ln(5x)+\frac{\sin(x)}{x}$.

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2) $(\sqrt{\cos(\sqrt{x})})'$.

Let $p(x)=f(g(h(x)))$, then $p'(x)=f'(x)\cdot g'(x)\cdot h'(x)$.

$f(x)=\sqrt{u}, u=\cos(\sqrt{x})\implies f'(x)=\frac{1}{2}\cdot\frac{1}{\sqrt{u}}=\frac{1}{2}\cdot\frac{1}{\sqrt{\cos(\sqrt{x})}}$,

$g(x)=\cos(v),v=\sqrt{x}\implies g'(x)=-\sin(v)=-\sin(\sqrt{x})$,

$h(x)=\sqrt{x}, h'(x)=\frac{1}{2}\cdot\frac{1}{\sqrt{x}}$.

$(\sqrt{\cos(\sqrt{x})})'=\frac{1}{2}\cdot\frac{1}{\sqrt{\cos(\sqrt{x})}}\cdot(-\sin(\sqrt{x}))\cdot\frac{1}{2}\cdot\frac{1}{\sqrt{x}}=$

$=-\frac{1}{4}\cdot\frac{\sin(\sqrt{x})}{\sqrt{x}\cdot\sqrt{\cos(\sqrt{x})}}$.

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3) $(e^{\ln(\tan(5x))})'$.

Let $p(x)=s(f(g(h(x))))\implies p'(x)=s'(x)\cdot f'(x)\cdot g'(x)\cdot h'(x)$.

$s(x)=e^{u}, u=\ln(\tan(5\cdot x))\implies s'(x)=e^{u}=e^{\ln(\tan(5\cdot x))}$,

$f(x)=\ln(v), v=\tan(5\cdot x)\implies f'(x)=\frac{1}{v}=\frac{1}{\tan(5\cdot x)}$,

$g(x)=\tan(w), w=5\cdot x\implies g'(x)=\frac{1}{\cos^{2}(w)}=\frac{1}{\cos^2(5\cdot x)}$,

$h(x)=5\cdot x\implies h'(x)=5$.

$(e^{\ln(\tan(5x))})'=e^{\ln(\tan(5x))}\cdot\frac{1}{tan(5x)}\cdot\frac{1}{\cos^{2}(5x)}\cdot5=$

$=5\cdot\tan(5x)\cdot\frac{1}{\tan(5x)}\cdot\frac{1}{\cos^{2}(5x)}=5\sec^{2}(x)$.

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4) $(\sin(2(\ln(\sec(x)))))'$.

$(\sin(2(\ln(\sec(x)))))'=\cos(2(\ln(\sec(x)))))\cdot2\cdot\frac{1}{\sec(x)}\cdot\frac{\sin(x)}{\cos^2(x)}=$

$=2\cdot\cos(2(\ln(\sec(x)))))\cdot\tan(x)$.

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5) $(\ln(\tan(x)))'=\frac{1}{\tan(x)}\cdot\frac{1}{\cos^2(x)}=\frac{1}{\sin(x)}=\csc(x)$.