用implicit differentiation求得:y’ = 1/cos y 精彩部分:进一步转化为:y’ = 1/√[1-(sin y)^2] 而sin y = x,因此,(sin y)^2 = x^2, 解出:y’ = 1/√(1-x^2) 书本内容: 第一,书本对反函数进行了进一步的解释(补充高中知识欠缺的我)。 Theorem 5.6 Reflec
Here they are: y = f(x) dy/dx 1 xn nxn-1 2 ex ex 3 ekx kekx 4 ax ax.ln a 5 ln x 1/x 6 log ax 1/x ln a 7 sin x cos x 8 cos x - sin x 9 tan x sec2 x 10 cot x - cosec2 x 11 sec x sec x.tan x 12 cosec x - cosec x. cot x 13 sinh x cosh x...
Example 4:Find the slope of the tangent line to the curvey= sinxat the point (π/2,1) Because the slope of the tangent line to a curve is the derivative, you find thaty′= cosx; hence, at (π/2,1),y′ = cos π/2 = 0, and the tangent line has a slope 0 at the point ...
f(x+Δx)−f(x)Δx=sinΔx2Δx2cos(x+Δx2)⇒limΔx→0sinΔx2Δx2cos(x+Δx2)=1×cosx=cosxf(x+Δx)−f(x)Δx=sinΔx2Δx2cos(x+Δx2)⇒limΔx→0sinΔx2Δx2cos(x+Δx2)=1×cosx=cosx[Since cos x is continuous and ...
(−a sin(a x)−a cos(a x)a cos(a x)−a sin(a x)) You can also perform differentiation of a vector function with respect to a vector argument. Consider the transformation from Cartesian coordinates(x,y,z)to spherical coordinates(r,λ,φ)as given by ...
Polar coordinates are used when data concerning position are received in the form of distance and direction, or, as the armed services would say, range and bearing. An instrument such as a radar scanner automatically measures r and θ. The definitions of cos θ and sin θ presented in the ...
TheGroupwillcontinuetoimplementthestrategiesofdifferentiationandcostleadershipinthesecond halfoftheyear,seekingtopursuestabledevelopmentanddeliversoundresultsbyenhancingprojectbasedoperationsandextendingitsadvantagesincostandtechnologyaswellasitsabilityinone-stop project delivery. ...
g_{2}'(0)=\lim_{x \rightarrow 0}{\frac{ g_{2}(x)}{x}}=\lim_{x \rightarrow 0}x\sin\frac{1}{x}=0 ,但在0以外的点,导数公式为cos(1/x) + 2x sin(1/x),这意味着在0的时候, g'_{2} 并不连续。所以g2的性质是这样的:连续、可微,但其导数在0不连续。 接下来我们试图用一...
f. y′ = 2xcos(x2) g. y= (sinx)2⟹y′ = 2 sinxcosx= sin 2x h. y′ = etanxsec2x i. Example 2: What is the equation of the tangent line to the curvey=exInxat the point (1, 0)? The first step is to find the slope of the tangent line atx= 1, which is the value...
assert np.allclose(grad(SIN())(x), np.cos(x)) v, g = value_and_grad(SIN())(x) # ...