2014年8月18日 · The derivative of tanx is sec^2x. To see why, you'll need to know a few results. First, you need to know that the derivative of sinx is cosx. Here's a proof of that result from first principles: Once you know this, it also implies that the derivative of cosx is -sinx (which you'll also need later). You need to know one more thing, which is …

2018年7月6日 · By definition: #d/dx tanx = lim_(h->0) (tan(x+h)-tanx)/h# Using the trigonometric formulas for the sum of two angles:

2015年10月10日 · Domain: (theta|theta!=kpi/2, where k is an odd integer) Range: (-oo,oo) Remember that tan=sin/cos therefore, you will have a vertical asymptope whenever cos=0. Cos=0 every odd multiple of pi/2. If you plug y=tan(x) into a graphing calculator you will see that the ends of each section continue on infinitely along the y-axis.

From the derivative of \\sin(x), \\cos(x) and \\tan(x) can be determined. \\cos(x) can be found by using the chain rule and the identity \\cos(x)=\\sin(x+90). \\tan(x ...

2015年9月21日 · 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 ...

If tanx =3/4, π < x <3 π /2, find the values of sin (x/2), cos (x/2) and tan (x/2)? Question #07cd9.

2016年11月4日 · dy/dx = sec^2x By definition If y=f(x) then: dy/dx=f'(x) =lim_(hrarr0)( (f(x+h)-f(x))/h ) So, with y=tanx we have: dy/dx = lim_(h rarr0)((tan(x+h)-tanx)/h) Using the ...

2016年12月9日 · Use logarithmic differentiation. Use the natural logarithm on both sides: ln(y) = ln(ln(x)^tan(x)) Use the property of logarithms ln(a^b) = (b)ln(a) ln(y) = tan(x)ln ...

2015年1月29日 · The period of y=tanx is pi. In fact: graph{tanx [-10, 10, -5, 5]}. The "fundamental" branch is in (-pi/2,pi/2

2017年7月28日 · y = -2cos^2(x)+3cos(x) The given equation, y'+ytan(x)=sin(2x) , is of the form: y' + P(x)y = Q(x) Where P(x) = tan(x), and Q(x) = sin(2x) It is known that the integrating factor is: mu(x) = e^(intP(x)dx) inttan(x)dx = log(sec(x)) mu(x) = e^log(sec(x)) = sec(x) Multiply the given equation by mu(x): y'sec(x)+tan(x)sec(x)y=sin(2x)/sec(x) We know that the left side integrates to mu(x)y and we are ...