Question

(iii) a2x log (sin √x)​

Answer 1

Step-by-step explanation:

Before answering it I assume that you Know following things

d/dx(f(g(x))) = d/dx(f(g(x))) * d/dx(g(x))

d/dx(lnx) = 1/x

d/dx (cot x) = -cosec^2 x

d/dx ( cosec x) = -cot x * cosec x

Now we have to calculate d/dx (log (sin x))

d/dx (log (sin x)) = d/dx (log (sin x)) * d/dx(sin x) * 1/ ln 10

=> d/dx (log (sin x)) = (1/sin x)*(cos x)* 1/ln 10= (cos x /sin x)*(1/ln 10)= cot x/ln 10

Now for second derivative we have to differentiate the 1st derivative

2nd derivative = d/dx (1st derivative)

2nd derivative = d/dx (cot x/ln 10) = -cosec^2 x/(ln 10)

Now for third derivative we have to differentiate 2nd derivative

3rd derivative = d/dx (-cosec^2 x/ln 10 )

= (-2*cosec x) * d/dx ( cosec x)*(1/ln 10)

= (- 2 cosec x) * ( -cot x * cosec x)*1/(ln 10)

= 2 * cosec^2 x * cot x* (1/ln 10)

Feel free to ask any doubt in comment section.

Hope it helps!!! Have a good day !

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