2014年8月22日 · You can find this derivative by applying the Chain Rule, with cosx as the inner function, and lnx as the outer function. Process: To apply the chain rule, we first find the derivative of the outer function, lnu, with u = cosx. Remember that the derivative of lnu = 1/u = 1/cosx. Now we just need to find the derivative of the inner function, cosx, and multiply it …

2016年7月13日 · For this particular, we'd have to use logarithmic differentiation, which works as follows: Let y = (lnx)cosx Taking the natural log (ln) of both sides yields lny = ln((lnx)cosx) lny = cosx ⋅ ln(lnx) Since the next step is to take derivatives, the rules we're going to use is d dx [lnu] = u' u Differentiating both sides gives y' y = −sinx ⋅ ln(lnx) + cosx ⋅ (1 x) ⋅ (1 lnx) (lnx 1) ⋅ ...

2014年8月10日 · In f(x) = ln(cos(x)), we have a function of a function (it's not multiplication, just sayin'), so we need to use the chain rule for derivatives: d/dx(f(g(x)) = f'(g(x ...

2015年10月16日 · -tanx Rule : d/dxlnu (x)=1/u* (du)/dx therefore d/dx ln (cosx)=1/cosx*d/ (dx)cosx =1/cosx*-sinx =-sinx/cosx =-tanx

2015年10月20日 · f (x) = ln (cos^2x) = ln ( (cosx)^2) = 2ln (cosx) f' (x) = 2 1/cosx (-sinx) " " (Chain rule: d/dx (lnu) = 1/u (du)/dx) = -2tanx

2016年12月20日 · Given that: eiθ = cosθ + isinθ cos(− θ) = cos(θ) sin(− θ) = −sin(θ) we can deduce that: cos(θ) = eiθ + e−iθ 2 So: cos(ln(x)) = eilnx + e−ilnx 2 ...

2016年11月1日 · How do you find the derivative of y = ln(cos x2)?

2015年8月30日 · Looking at ∫cos(lnx)dx, we realize that if can't integrate straight away. Next thought is, perhaps we could use substitution. We would need cos(lnx) 1 x to integrate by substitution. Running out of ideas, let's try putting in the "missing" 1 x and an x to make up for it and try integrating by parts. (Yes, really.

2018年5月29日 · We use a technique called logarithmic differentiation to differentiate this kind of function. In short, we let y = (cos (x))^x, Then, ln (y) = ln ( (cos (x))^x) ln (y ...

Let y = (lnx)cosx, so that ln y= cos x ln ln x Now differentiate both sides, 1 y dy dx = − sinxlnlnx + cosx 1 lnx 1 x dy dx = (lnx)cosx(−sinxlnlnx + cosx xlnx)