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  • Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10 point 1 question 5 maths textbook solution

Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 5 maths textbook solution

Answers (1)

Answer:

\frac{e\sqrt{2x}}{\sqrt{2x}}

Hint:

Use first principle formula to find the differentiation

Given:

e\sqrt{2x}

Solution:

Let,

f\left ( x \right )=e^{\sqrt{2x}}
f\left ( x+h \right )=e^{\sqrt{2\left ( x+h \right )}}

Now, we will use the formula of first principle to find the differentiation

\frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}

\frac{d}{dx}\left ( e^{\sqrt{2x}} \right )=\lim_{h\rightarrow 0}\frac{e^{\sqrt{2\left ( x+h \right )}}-e^{\sqrt{2x}}}{h}

Take 2e^{\sqrt{2x}} common from numerator

                     = \lim_{h\rightarrow 0}\frac{e^{\sqrt{2x}}\left [ \left ( \frac{e^{\sqrt{2\left ( x+h \right )}}}{e^{\sqrt{2x}}} \right )-1 \right ]}{h}

                     =\lim_{h\rightarrow 0}\cdot e^{\sqrt{2x}}\frac{\left [ \left ( \left [ e^{\sqrt{2\left ( x+h \right )}} \right ] \right ) -1\right ]}{h}                        \left [ \because \frac{a^{m}}{a^{n}} =a^{m-n}\right ]

 

Multiply and divide by \left ( \sqrt{2\left ( x+h \right )}-\sqrt{2x} \right )

                     = \lim_{h\rightarrow 0}e^{\sqrt{2x}}\times \frac{\left [ \left ( e^{^{\sqrt{2\sqrt{\left ( x+h \right )}-\sqrt{2x}}}} \right ) -1\right ]}{\left ( \sqrt{2\left ( x+h \right )}-\sqrt{2x} \right )}\times \left [ \frac{\sqrt{2\left ( x+h \right )-\sqrt{2x}}}{h} \right ]

 

                     = \lim_{h\rightarrow 0}e^{\sqrt{2x}}\times 1\times \left [ \frac{\sqrt{2\left ( x+h\right )}-\sqrt{2x}}{h} \right ]                                            \left [ \because \lim_{h\rightarrow 0}\frac{e^{x}-1}{x} =1\right ]

 

                     =\lim_{h\rightarrow 0}e^{\sqrt{2x}}\times \left [ \frac{\sqrt{2\left ( x+h \right )-\sqrt{2x}}}{h} \right ]

Now, e^{\sqrt{2x}} is a function of x and unit applied on variable ‘h’. So,e^{\sqrt{2x}}  can be taken outside became w.r.t variable ‘h’, x will remain constant

= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\left [ \frac{\sqrt{2\left ( x+h \right )}-\sqrt{2x}}{h} \right ]

    

Rationalising the numerator

                  = e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\left [ \frac{\sqrt{2\left ( x+h \right )}-\sqrt{2x}}{h}\times \frac{\sqrt{2\left ( x+h \right )}+\sqrt{2x}}{\sqrt{2\left ( x+h \right )}+\sqrt{2x}} \right ]

                   = e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\left [ \frac{\left ( \sqrt{2\left ( x+h \right )} \right )^{2}-\left ( \sqrt{2x} \right )^{2}}{h\times \left ( \sqrt{2\left ( x+h \right )} \right )} \right ]              \left[\because(a+b)(a-b)=a^{2}-b^{2}\right]

                   = e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2\left ( x+h \right )-2x}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}

                   = e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2x+2h-2x}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}     

                 = e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2h}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}

                 = e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}

                  = e^{\sqrt{2x}}\times \frac{2}{2\sqrt{2x}}

                  =\frac{ e^{\sqrt{2x}}}{\sqrt{2x}} 

Hence, the differentiation of e^{\sqrt{2x}} is\frac{ e^{\sqrt{2x}}}{\sqrt{2x}}

             

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