2015年7月28日 · d/dx (2^x) = 2^x * ln2 In order to be able to calculate the derivative of 2^x, you're going to need to use two things the fact that d/dx(e^x) = e^x the chain rule The idea here is that you can use the fact that you know what the derivative of e^x is to try and determine what the derivative of another constant raised to the power of x, in this case equal to 2, is. To do that, …

Using first principles, the derivative of the exponential function c^x can be simplified, however, determining the actual limit is best done by using a computer.

2016年1月11日 · First, note that bx = eln(bx) = exlnb This allows us to differentiate the function using the chain rule: d dx [exlnb] = exlnb ⋅ d dx [xlnb] Just like d dx [5x] = 5, d dx [xlnb] = lnb, since lnb will always be a constant. This gives us a derivative of: exlnb ⋅ lnb Now, recall that exlnb = bx. This gives us our final, differentiated result:

2015年6月1日 · Use the Chain Rule: d dx (ln(sin(x))) = 1 sin(x) ⋅ cos(x) = cos(x) sin(x) = cot(x)

2018年3月20日 · d dx lnf (x) = f '(x) f (x) ⇒ d dx (ln(lnx)) = d dx(lnx) lnx = 1 x lnx 1 xlnx

2016年1月19日 · 5cos5x Use the chain rule. The chain rule states that, in the case of a sine function, d/dx[sinu]=cosu*(du)/dx More generally, the chain rule says to identify an inside function and an outside function. Here, the outside function is sinx, and the inside function is 5x. The chain rule then says to differentiate the outside function, and the derivative of sinx is cosx. With this …

2016年11月10日 · d/dx (1/sinx)= -cotx cscx There are several methods to do this: Let y= 1/sinx (=cscx) Method 1 - Chain Rule Rearrange as y= (sinx)^-1 and use the chain rule: { ("Let ...

2016年7月23日 · (A) here f (g (x))=sin^2 (2x)= (sin2x)^2 rArrf' (g (x))=2sin2x Note, 2 applications of color (blue)"chain rule" required for g' (x) g (x)=sin2xrArrg' (x)=cos2x.d/dx (2x)=2cos2x "------------------------------------------------------------------" Substitute these values into (A) 2sin2x.2cos2x=4sin2xcos2x color (orange)"Reminder" color (red ...

2017年10月22日 · STEP 1: secx = 1 cosx STEP 2: Use the quotient rule to find the derivative of secx = 1 cosx d dx1 ⋅ (cosx) − d dxcosx ⋅ (1) (cosx)2 STEP 3: Simplify = (0)cosx − (− sinx) (cosx)2 = sinx (cosx)(cosx) = (sinx cosx)(1 cosx) = tanxsecx

2016年11月3日 · 2xy + 2y2 = 13 Differentiating wrt x and applying the product rule gives us: 2{(x)(dy dx) + (1)(y)} +4y dy dx = 0 x dy dx + y + 2y dy dx = 0 ⇒ dy dx = − y x + 2y Differentiating again wrt x and applying the product rule (twice) gives us: ∴ {(x)(d2y dx2) + (1)(dy dx)} + dy dx + 2{(y)(d2y dx2) + (2 dy dx)(dy dx)} = 0 ∴ x d2y dx2 + dy dx + dy dx + 2y d2y dx2 + 2(dy dx)2 = 0 …