
What is the derivative of 2^x? | Socratic
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 …
From First Principles - Calculus | Socratic
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.
What is the derivative of b^x where b is a constant? | Socratic
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, …
What is the derivative of ln(sinx)? | Socratic
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)
What is the derivative of y=ln(ln(x))? | Socratic
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
What is the derivative of sin 5x? | Socratic
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 …
How do you find the derivative of 1/sinx? | Socratic
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 ...
How do you differentiate sin^2(2x)? | Socratic
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 "------ …
Hw do you prove that the d/dx (secx) = sec x tan x - Socratic
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) …
How do you find ( (d^2)y)/ (dx^2)? - Socratic
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 …