Question

differentiate it......plzzzz​

Answer 1

Answer:

Step-by-step explanation:

soving these by chain rule

e^(cosec x)^2

first differentiate the first term

which is same

e^(cosec x)^2 *differentiation of cosec x^2

e^(cosec x)^2*2(cosec x)*differentiation of cosec x

final answer is

e^(cosec x)^2*2(cosec x)*-cosec x*cot x

2)

e^log(log x)

differentiating with respect to x

e^log(log x)*multiplied by differentiation of log(log x)

e^log(log x)*1/log x*multiplied by differentiation of log x

e^log(log x)*1/log x*1/x

3)a^log(1+log x)

let us assume this as y

y=a^log(1+log x)

take log on both sides

log y=log(1+log x)*log a

differentiating with respect to x

1/y dy/dx=loga(1/1+log x*1/x)

dy/dx=y(loga(1/1+log x*1/x))

dy/dx=a^log(1+log x)(loga(1/1+log x*1/x))

4)(a^root x) ^sin x

let us assume above as y

y=(a^root x) ^sin x

take log on both sides

log y=sin x log(a^root x)

differentiating with respect to x

1/y dy/dx=cos x* log(a^root x)+sin x(1/a root x)(0)

1/y dy/dx=cos x* log(a^root x)

dy/dx=cos x* log(a^root x)*y

dy/dx=cos x* log(a^root x)*(a^root x) ^sin x