What is the derivative of #y=ln(cos^2ɵ)#? - Socratic
Jul 31, 2015 · Assuming that we want y' = dy/(d theta), in simplest form: y' = -2tan theta y = ln(cos^2 theta) Method 1 Leave it as is and use the chain rule twice: y' = 1/cos^2 theta * d/(d theta) (cos^2 theta) = 1/cos^2 theta * 2cos theta * d/(d theta) (cos theta) = 1/cos^2 theta * 2cos theta * d/(d theta) (cos theta) = 1/cos^2 theta * 2cos theta * (-sin theta) = -2 sintheta/costheta = -2tan theta Method 2 ...
What is the derivative of #ln(cos 2x)#? - Socratic
Jan 25, 2016 · Here I would use the Chain Rule applied to your function: #f(x)=ln(cos(2x))# deriving #ln# as it is, multiplying by the derivative of #cos# as it is (red) and finally multiplying by the derivative of #2x# (blue):
What is the derivative of (ln x) ^ cos x? | Socratic
Let #y=(ln x)^cos x#, so that ln y= cos x ln ln x. Now differentiate both sides, #1/y dy/dx = -sinx ln ln x + cos x 1/lnx 1/x#
How do you differentiate f(x)=ln(cos(e^(x) )) using the ... - Socratic
Mar 17, 2016 · How do you differentiate #f(x)=ln(cos(e^(x) )) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule. 1 Answer
How do you find dy/dx given #y=ln(cos x)#? - Socratic
dy/dx=-tanx Given: y=ln(cosx) Let t=cosx Then, y=ln(t) By chain rule dy/dx=dy/dt.dt/dx y=lnt dy/dt=1/t t=cosx dt/dx=-sinx Thus dy/dx=(1/t).(-sinx) t=cosx dy/dx=(1 ...
What is the derivative of #(sinx)^(cos^(-1)x)#? - Socratic
Jul 4, 2017 · dy/dx = (sinx)^(cos^(-1)x) \\ (cos^(-1)x \\ cotx -(ln sinx)/sqrt(1-x^2) ) Let: y=(sinx)^(cos^(-1)x) Take Natural logarithms of both sides: ln y = ln {(sinx)^(cos^(-1 ...
How do you find the arc length of the curve y=ln(cosx) over the ...
Feb 25, 2015 · The answer is: #ln(sqrt2+1)#. The lenght of a function written in cartesian coordinates is: #L=int_a^bsqrt(1+[f'(x)]^2)dx#.
How do you differentiate y =sin(ln(cos x)) using the chain rule ...
Jan 15, 2016 · Differentiate using the 'chain rule' : #dy/dx = cos(ln(cosx)) .d/dx (ln(cosx)) # #dy/dx = cos(ln(cosx)).( 1/cosx . d/dx(cosx)) #
How do I find the derivative of y = sin^2 x + cos^2x + ln(e
Jan 27, 2015 · Derivatives of y=sin^2x+cos^2x+ln(e^x) can be easily find by finding the derivative for each component as the derivative of the sum is the sum of the derivative. Derivative of sin^2x is 2sin(x) * d(sin(x)) = 2sin(x)cos(x) Derivative of cos^2x is 2cos(x) * d(cos(x)) = -2cos(x)sin(x) Derivative of ln(e^x) is 1/e^x * d(e^x) = 1/e^x*e^x = 1 [*] …
What is the arc length of #f(x)=cosx# on #x in [0,pi]#? - Socratic
Jun 1, 2018 · 2E(-1), where E is the complete elliptic integral of the second kind. There is no simpler way of expressing this analytically. Numerically, this is approximately equal to 3.82. Arc length of a curve f(x) is given by the formula s=int_(x_0)^(x_1) sqrt(1+f'(x)^2) dx. Note that it is often a difficult problem to calculate this integral, and many classes of …