Question

Differentiate w.r.t.x: 3cos (log x)​

Answer 1

Answer:

Solution

y=3cos(logx)+4sin(logx)y=3cos(logx)+4sin(logx)

Let logx=t⇒1xdx=dt⇒dtdx=1xlogx=t⇒1xdx=dt⇒dtdx=1x

Then, y=3cost+4sinty=3cost+4sint

⇒dydt=−3sint+4cost⇒dydt=-3sint+4cost

∴dydx=dydt⋅dtdx=(−3sint+4cost)1x∴dydx=dydt⋅dtdx=(-3sint+4cost)1x

⇒xdydx=4cost−3sint⇒xdydx=4cost-3sint

Differentiating w.r.t. xx,

⇒xd2ydx2+dydx=(−4sint−3cost)dtdx⇒xd2ydx2+dydx=(-4sint-3cost)dtdx

⇒xd2ydx2+dydx=−

hope it will help you

Answer 2

Step-by-step explanation:

d/dx cosx=-sinx

d/dx logx= 1/x

3(-sin(logx) ) ×d/dx logx

-3sin(logx)×1/xAnswer