graph{tan x [-10, 10, -5, 5]}. See explanation. graph{tan x [-10, 10, -5, 5]} The number of lines you have to draw depends on the domain provided. If the domain is not specifically provided, you will have to draw them all (that's what the function does). If you're doing it by hand, I would first trace some vertical construction lines (asymptotic lines) at pi/2, -pi/2, etc..., where tan x is ...

2015年3月27日 · How do you use logarithmic differentiation to find the derivative of #y=(tanx)^(1/x)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer

2016年8月12日 · Take the natural logarithm of both sides. ln[y] = ln[(lnx)^(tanx)] Simplify the right-hand side using the rule ln[a^n} = nlna: ln[y] = tanx[ln(lnx)] Differentiate both sides. 1/y(dy/dx) = square Inset: square The derivative of the right hand side is fairly complex. We have to first find the derivative of ln(lnx) using the chain rule, followed by the derivative of tanx xx ln(lnx), using the ...

How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x)

2017年1月29日 · How do you find the derivative of #y = [(tanx - 1) / secx]#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer

2017年9月1日 · The area is 2(1 -ln2) We start by solving the equation tanx = 2sinx This implies that tanx - 2sinx =0 Or sinx/cosx - 2sinx = 0 Putting on a common denominator, we can say (sinx - 2sinxcosx)/cosx = 0 The only possible solutions to this equation will be sinx - 2sinxcosx = 0 sinx(1 -2cosx) = 0 sinx = 0 or cosx = 1/2 x = 0, pi, pi/3, (5pi)/3 By …

2015年2月16日 · The period of the function #y=tanx# is #pi#, it can be seen from the graph: graph{tanx [-5, 5, -2.5, 2.5]} The fundamental branche is in #(-pi/2,pi/2)#. The #4# doesn't change the period, but only the shape, more steep, as it is possible to see: graph{4tanx [-5.625, 5.625, …

2018年4月18日 · Here, #y=(secx+tanx)(secx-tanx)# Using product rule, #(dy)/(dx)=(secx+tanx)d/(dx)((secx-tanx))+(secx-tanx)d/(dx) ((secx+tanx)#

2015年11月10日 · It can't be done As y=arctanx is the inverse of y=tanx it can't be done. As we know tan goes on forever but when working in degrees you can never input values of 90, 270 and so on. This makes it impossible to define a domain for y=tanx as you'd have an infinite number of x != 90 and so on exceptions. Additonaly the range of tan goes on into infinity which makes it even more tricky. The range ...

2018年4月27日 · 0 By trigonometric identities: cotA=cosA/sinA=1/tanA Therefore, y=tanx(1/tanx)=tanx/tanx=1 So, dy/dx=0 You could try and use product rule as well as the quotient rule for 1/tanx and this should give you the same result of 0. Hope this helps!