2015年6月1日 · What is the derivative of #ln(sinx)#? Calculus Differentiating Trigonometric Functions Differentiating sin ...

2018年4月22日 · #"differentiate using the "color(blue)"chain rule"# #"Given "f(x)=g(h(x))" then"# #f'(x)=g'(h(x))xxh'(x)larrcolor(blue)"chain rule"#

2018年5月19日 · cotx We use the chain rule, which states that, dy/dx=dy/(du)*(du)/dx Let u=sinx,:.(du)/dx=cosx. Then, y=lnu,dy/(du)=1/u.

2016年10月30日 · #y=ln((sinx)^2)# Which means that: #e^y=(sinx)^2# Use implicit differentiation on the left hand side of the equation and the chain rule on the right hand side of the equation: #e^y*(dy)/(dx)=2sinx*cosx# Divide expressions on both sides of the equation by #e^y#: #(dy)/(dx)=(2sinx*cosx)/e^y# Don't forget that #e^y# is #(sinx)^2#:

2016年12月2日 · int_0^(pi/2) ln(sin x)dx = pi/2ln(1/2) Suppose it is convergent and put: S = int_0^(pi/2) ln(sin x)dx Substitute t=pi/2-x S=-int_(pi/2)^0ln(sin(pi/2-t))dt = int_0^(pi/2) ln( cos t )dt Summing the two expressions: 2S = int_0^(pi/2) ln(sin x)dx + int_0^(pi/2) ln( cos x )dx and as the integral is linear: 2S = int_0^(pi/2) ln(sin x)dx + ln( cos …

2016年1月9日 · Let #y=(ln(x))^sin(x)#. To find the derivative of such a problem we need to take logarithms on both the sides. #ln(y)=ln((ln(x))^sin(x))#

2016年11月6日 · Use implicit differentiation along with the chain and product rules. We can find the derivative of this function implicitly. In other words, we will find the derivative of y, which will then allow us to find the derivative of sin(x)^(ln(x)). First, we want to get rid of the lnx exponent. We can do that by taking the natural log of both sides and …

2017年9月25日 · #"differentiate using the "color(blue)"chain rule"# #"given "y=f(g(x))" then "# #dy/dx=f'(g(x))xxg'(x)larr" chain rule"#

2018年5月27日 · lim_(xto0)ln(sinx)=-oo lim_(xto0)ln(sinx) Set sinx=y x->0 y->0 =lim_(yto0)lny=-oo

2018年1月30日 · # ln y = ln {(sinx)^(sinx) }# And using the properties of logarithms we have: # ln y = sinx \ ln sinx# We can now readily differentiate wrt #x# by applying the chain rule (or implicit differentiation the LHS and the chain rule and the product rule on the RHS: # 1/y \ dy/dx = (sinx)(1/sinx cosx) + (cosx)ln sinx # Which we can simplify: