Differentiate w.r.t.x: 3cos (log x)
Answers
Answered by
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
Answered by
2
Step-by-step explanation:
d/dx cosx=-sinx
d/dx logx= 1/x
3(-sin(logx) ) ×d/dx logx
-3sin(logx)×1/xAnswer
Similar questions