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For ln x 2, both positive and negative values of x are allowed because x is squared. The only constraint on x is that x ≠ 0. The function domain includes negative x. The derivative is the slope ...
1 Answer. Anjali G. Mar 19, 2017. dy dx = 2lnx x. Explanation: y = (lnx)2. Use the Chain Rule: dy dx = 2(lnx)(1 x) dy dx = 2lnx x.
The corresponding expression is therefore: $$ b (n, k) = \frac {1} {2} [1 + (-1)^ {n+1}] + 2k $$. The exponents of $ x^2 + x - 2 $ in each term's denominator ($ c $) vary according to $ k $ (starting with $ k = 0 $) following the sequence of consecutive numbers which have their initial value determined by $ n $ according to the sequence {1, 1 ...
Explanation: Just to show the versatility of calculus, we can solve this problem through implicit differentiation. Raise both side to the power of e. y = ln(x2) ey = eln(x2) ey = x2. Differentiate both sides with respect to x. The left side will require the chain rule. ey(dy dx) = 2x.
Explanation: We can use chain rule here. We can write f (x) = ln(lnx2) as. f (x) = ln(g(x)), g(x) = ln(h(x)) and h(x) = x2. then df dg = 1 g(x), dg dh = 1 h(x) and dg dh = 2x. and using chain rule as df dx = df dg × dg dh × dh dx. = 1 g(x) × 1 h(x) × 2x. = 1 lnx2 ⋅ 1 x2 ⋅ 2x. = 2 xlnx2.
We have: y = ln(x^2+y^2) Method 1: Implicit differentiation, as is: Using the chain rule: dy/dx = 1/(x^2+y^2)(2x+2ydy/dx) " " = (2x)/(x^2+y^2) + (2y)/(x^2+y^2)dy/dx ...
Explanation: As f (x) = (lnx)2. First derivative f '(x) = 2lnx × 1 x = 2lnx x. and second derivative can now be found using quotient rule. f ''(x) = x × 2 x −2lnx ×1 x2 = 2(1 − lnx) x2. Answer link.
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(2x)/(x^2+1) You have a composed function f(g(x)), where f(x)=ln(x), and g(x)=x^2+1 The rule for deriving composite functions is d/dx f(g(x)) = f'(g(x))*g'(x) which can be translated as "compute the derivative of the outer function with the inner function as argument, and multiply the derivative of the inner function". To complete our scheme, we need the derivatives: we have f(x)=ln(x ...
1 Answer. use the quotient rule, which should be memorised. f (x) = u v ⇒ f '(x) = vu' −uv' v2. f (x) = lnx x2. u = lnx ⇒ u' = 1 x. v = x2 ⇒ v' = 2x. f '(x) = x2 × 1 x − 2xlnx (x2)2. f '(x) = x −2xlnx x4 = x(1 −2lnx) x4 = x(1 − lnx2) x4 x3.