D-d x-ee x-sin b x-c-

The correct option is C =e^ax[a sin(bx+c) b cos(bx+c)]/sin^2(bx+c)d/dx ( e^ax/sin(bx+c) )=ae^axsin(bx+c) be^axcos(bx+c)/{ sin(bx+c) }^2 =e^ax[a sin(bx+c) b co ...

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Byju's Answer

Standard XII

Mathematics

Quotient Rule of Differentiation

d/d x[ee x/si...

Question

$\frac{d}{d x} [\frac{e^{a x}}{s i n (b x + c)}] =$

= \frac{e^{a x} [a s i n (b x + c) + b c o s (b x + c)]}{s i n^{2} (b x + c)}

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= \frac{e^{a x} [a s i n (b x + c) - b c o s (b x + c)]}{s i n (b x + c)}

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= \frac{e^{a x} [a s i n (b x + c) - b c o s (b x + c)]}{s i n^{2} (b x + c)}

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Solution

The correct option is C $= \frac{e^{a x} [a s i n (b x + c) - b c o s (b x + c)]}{s i n^{2} (b x + c)}$
$\frac{d}{d x} (\frac{e^{a x}}{s i n (b x + c)})$ $= \frac{a e^{a x} s i n (b x + c) - b e^{a x} c o s (b x + c)}{{s i n (b x + c)}^{2}} = \frac{e^{a x} [a s i n (b x + c) - b c o s (b x + c)]}{s i n^{2} (b x + c)}$

Suggest Corrections

ddx[eaxsin(bx+c)]=

The correct option is C =eax[a sin(bx+c)−b cos(bx+c)]sin2(bx+c)ddx(eaxsin(bx+c))=aeaxsin(bx+c)−beaxcos(bx+c){sin(bx+c)}2=eax[a sin(bx+c)−b cos(bx+c)]sin2(bx+c)

$\frac{d}{d x} [\frac{e^{a x}}{s i n (b x + c)}] =$