Differentiate tan^(-1) ((3a^2x-x^3)/(a^3-3a x^2)), View Solution Differentiatetan−1(3x−x31−3x2),|x|<1√3w.r.ttan−1(x√1−x2) View Solution Differentiatetan−1(3a2x−x3a3−3ax2) View Solution Differentiatetan−1(4√x1−4x) ...
f(x) = tan^-1 [ (3x-x^3)/(1-3x^2)] let x= tan theta theta= tan^-1 x so f(x) = tan^-1 [ tan 3 theta] x > 1/sqrt3 f(x) = tan^-1 (tan 3 theta) 1/sqrt3 < tan theta < oo pi/6 < theta < pi/2 pi/2 < 3 theta < 3 pi/2 f(x) = tan^-1 (-tan
Find {dy} / {dx} by implicit differentiation. tan^{-1} (x^2 y) = x + x y^2 Differentiate y= 2x/ 5-tanx Differentiate and simplify the following: (a) (ln(8x) + e^(cos x))^10 (b) 1/2 sin^(-1) x/2-1/5cos^2(6x) ...
The implicit differentiation technique is given as: dydx=−[∂∂x[f(x,y)]∂∂y[f(x,y)]] Answer and Explanation: We have the following given data {eq}\begin{align} \ \tan^{-1}(xy) &= 3x^2 + \sin^{-1}(y)\[0.3cm] \frac{\text{d}y}{\text{d}x} &=...
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23K The rules of differentiation are useful to find solutions to standard differential equations. Identify the application of product rule, quotient rule, and chain rule to solving these equations through examples. Related to this Question Find Integral_{2}^{3} [8 f(x) - g(x)] dx if Inte...
Since differentiation and integration are inverse operations, we can make some strong statements about these two tasks. The Fundamental Theorem of Calculus makes some of these statements, one of which is below. {eq}\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x)) h'(x) - ...
adipocyte differentiationTEA domain transcription factor 1 ( TEAD1 ), also called TEF-1 , acts as a transcriptional enhancer to regulate muscle-specific gene expression. However, the role of TEAD1 in regulating intramuscular preadipocyte differentiation in goats is unclear. The aim of this study ...
Differentiation Interactive Applet - trigonometric functions. In words, we would say: The derivative ofsinxiscosx, The derivative ofcosxis−sinx(note the negative sign!) and The derivative oftanxissec2x. Now, ifu=f(x)is a function ofx, then by using the chain rule, we have: ...
在等式2nsinx2ncosx2ncosx2n−1⋯cosx2=sinx 两端加绝对值再取对数：ln...