Answer to Question #93604 in Calculus for Ezekiel

Question

1) Find dy/dx of the following using first principle
a) y = 1/√x
b) y = √x+1

2) Find the first derivative of the following :
a) y= e -1/2x +cosx sinx
b) y= √E ax sinx

3) Find dy/dx of the following :
a) sinx cosx/Cos2x
b) √x+1/ √x - 1

Answer 1

${y}'\left(x\right)=\lim\limits_{h\to 0}\frac{y\left(x+h\right)-y\left(x\right)}{h}$

a)

${y}'\left(x\right)=\lim\limits_{h\to 0}\frac{\frac{1}{\sqrt {x+h} } -\frac{1}{\sqrt x}}{h}$

Move the h  into the numerator functions' denominators

$=\lim\limits_{h\to 0}\frac{1}{h\sqrt {x+h} } -\frac{1}{h\sqrt x}$

Get common denominators by cross-multiplying.

$=\lim\limits_{h\to 0}\frac{1}{h\sqrt {x+h} } \frac{\sqrt {x}}{\sqrt {x} }-\frac{1}{h\sqrt x}\frac{\sqrt {x+h}}{\sqrt {x+h} }$

$=\lim\limits_{h\to 0}\frac{\sqrt {x}}{h\sqrt {x+h}\sqrt {x} }-\frac{\sqrt {x+h}}{h\sqrt {x+h}\sqrt x}$

Multiply by the complex conjugate of the resultant numerator as a fraction equal to 1

$=\lim\limits_{h\to 0}\frac{\sqrt {x}-\sqrt {x+h}} {h\sqrt {x(x+h)} } * \frac{\sqrt {x}+\sqrt {x+h}} {\sqrt {x}+\sqrt {x+h}}$

$=\lim\limits_{h\to 0}\frac{{x}- (x+h)} {h\sqrt {x(x+h)}(\sqrt {x}+\sqrt {x+h}) }$

Thus, you get a cancellation of x  in the numerator and h  is by itself.

$=\lim\limits_{h\to 0}\frac{-h} {h\sqrt {x(x+h)}(\sqrt {x}+\sqrt {x+h}) } =\\$

$=\lim\limits_{h\to 0}\frac{-1} {\sqrt {x(x+h)}(\sqrt {x}+\sqrt {x+h}) } =\frac{-1} {2x\sqrt {x} }=\frac{-1} {2x^{3/2 }}$ .

b)

${y}'\left(x\right)=\lim\limits_{h\to 0}\frac{\sqrt {x+h+1} -\sqrt {x+1} }{h}$

${y}'\left(x\right)=\lim\limits_{h\to 0}\frac{\sqrt {x+h+1} -\sqrt {x+1} }{h}=\\ =\lim\limits_{h\to 0}\frac{(\sqrt {x+h+1} -\sqrt {x+1}) (\sqrt {x+h+1} +\sqrt {x+1}) }{h (\sqrt {x+h+1} +\sqrt {x+1})}=\\ =\lim\limits_{h\to 0}=\frac{( {x+h+1} - x-1) }{h (\sqrt {x+h+1} +\sqrt {x+1})} = \\ =\lim\limits_{h\to 0}\frac{ {h} }{h (\sqrt {x+h+1} +\sqrt {x+1})}=\\=\lim\limits_{h\to 0}\frac{ {1} }{ (\sqrt {x+h+1} +\sqrt {x+1})}=\\ =\frac{ {1} }{ \sqrt {x+1} +\sqrt {x+1}}=\frac{ {1} }{ 2\sqrt {x+1}}.$

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2.

a)

$y= \exp(1/2x) +\cos x \sin x$

Find the first derivative

$y'= ( e^{1/2x} +\cos x \sin x)'= \\ = ( e^{1/2x}) + \frac{1}{2} \sin 2x)'=\\ = \frac{1}{2} e^{1/2x} + \frac{1}{2} 2\cos 2x= \\ = \frac{1}{2} e^{1/2x} +\cos 2x$

$y'= \frac{1}{2} e^{1/2x} +\cos 2x$ .

b)

$y= \sqrt{ e^{ax}} \sin x$

$y'= (y= \sqrt{ e^{ax}} \sin x)'= \\ = (\sqrt{ e^{ax}})' \sin x+ \sqrt{ e^{ax}}( \sin x)'=\\ = \frac{a e^{ax}\sin x} {2\sqrt{ e^{ax}}}+ \sqrt{ e^{ax}}( \cos x)= \frac{e^{ax}(a \sin x+ 2\cos x)} {2\sqrt{ e^{ax}}}$.

$y'= \frac{e^{ax}(a \sin x+ 2\cos x)} {2\sqrt{ e^{ax}}}$ .

3.

a)

$\frac{dy} {dx}= \left(\frac{\sin x \cos x}{\cos 2x} \right)'= \left(\frac{1}{2}\frac{\sin 2x }{\cos 2x} \right)'= \left( \frac{1}{2}\tan 2x \right)'= \\ \frac{1}{2}\frac{1}{\cos^2 2x} *2=\frac{1}{\cos^2 2x}$

$\frac{dy} {dx}= \frac{1}{\cos^2 2x}$ .

b)

$\frac{dy} {dx}= \left(\frac{ {\sqrt {x+1}} }{ \sqrt {x-1}} \right)'=\left(\frac{ {(\sqrt {x+1})'\sqrt {x-1}- (\sqrt {x-1})'\sqrt {x+1}} }{ (\sqrt {x-1})^2} \right)=\\ =\left(\frac{ {\frac{ {1} }{2 \sqrt {x+1}}\sqrt {x-1}- \frac{ {1} }{2 \sqrt {x-1}}\sqrt {x+1}} }{ {x-1}} \right)=\frac{ {1} }{ {x-1}}\frac{ {-2} }{ {2 \sqrt {x+1} \sqrt {x-1}}}= \\ =\frac{ {-1} }{ { \sqrt {x+1} ({x-1})^\frac{3}{2}}}$ =

$\frac{dy} {dx}= =\frac{ {-1} }{ { \sqrt {x+1} ({x-1})^\frac{3}{2}}}$ .