If {{{l}}₁ = {{{d}}}{{{{dx}}}}≤ft( {{{{e}}^{ {{x}}}}} ) {{{l}}₂ = { }_{{{h}} → 0} {{{{{e}}^{ ≤ft( {{{x}} + {{h}}} )}} - {{{e}}^{ {{x}}}}}}{{{h}}} {{{l}}₃ = {{{e}}^{ {{x}}}} {{x\;dx}} Then which one of the following is correct?

Correct Answer is : ddx(l3)=l2{{{d}}}{{{{dx}}}}≤ft( {{{{l}}₃} ) = {{{l}}₂dxd(l3)=l2

{{{d}}}{{{{dx}}}}≤ft( {{{{l}}₃} ) = {{{l}}₂

Correct Answer is : ddx(l3)=l2{{{d}}}{{{{dx}}}}≤ft( {{{{l}}₃} ) = {{{l}}₂dxd(l3)=l2

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 73 of 120

Marks: +1, -0

{{\rm{l}}_1} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{\sin {\rm{x}}}}} \right)

{{\rm{l}}_2} = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{{\rm{e}}^{\sin \left( {{\rm{x}} + {\rm{h}}} \right)}} - {{\rm{e}}^{\sin {\rm{x}}}}}}{{\rm{h}}}

{{\rm{l}}_3} = \smallint {{\rm{e}}^{\sin {\rm{x}}}}\cos {\rm{x\;dx}}

Then which one of the following is correct?

Solution:

Concept:

l_1

and

l_2

both represent the derivative of

e^{\sin x}

;

l_3

is the indefinite integral of

e^{\sin x}\cos x

, which equals

e^{\sin x}+c

Explanation:

Step 1: Compute

l_1

and

l_2

l_1 = \frac{d}{dx}\left(e^{\sin x}\right)

l_2 = \lim_{h\to 0}\frac{e^{\sin(x+h)}-e^{\sin x}}{h}

is the limit definition of the same derivative.

Hence

l_1 = l_2

Step 2: Compute

l_3

l_3 = \int e^{\sin x}\cos x\,dx

Let

\sin x = t

, then

\cos x\,dx = dt

l_3 = \int e^t dt = e^t + c = e^{\sin x} + c

Step 3: Differentiate

l_3

\frac{d}{dx}(l_3) = \frac{d}{dx}\left(e^{\sin x}+c\right) = \frac{d}{dx}\left(e^{\sin x}\right)

This equals

l_1

(and therefore

l_2

Thus

\frac{d}{dx}(l_3) = l_2

Answer:

Option B:

\frac{d}{dx}(l_3) = l_2

is correct.

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