Correct forms of Trigonometric Derivative Functions

• \(\frac{d(sinx)}{dx}=cosx\)
• \(\frac{d(cosx)}{dx}=-sinx\)
• \(\frac{d(secx)}{dx}=secxtanx\)
• \(\frac{d(tanx)}{dx}=sec^2 x\)

Put x = at : y = t
\(=lt_{t\rightarrow 0}\frac{sin(t)}{a^n.t^n}\)
\(=lt_{t\rightarrow 0}\frac{1}{a^nt^{n-1}}\frac{sin(t)}{t}\)
By varying n we get different limits
Hence, Does Not Exist is right answer.

\(\lim_{x\rightarrow 0}\frac{ax^3+b sin(x)+c cos(x)}{x^5}=1\)
Now expanding terms of sin(x) and cos(x) and rearranging in powers of x,x³ and x⁵ and so on,we get
=>\(\lim_{x\rightarrow 0}\frac{x(b+c)-x^3(\frac{b}{6}+\frac{c}{2}-a)+x^5(\frac{b}{120}+\frac{c}{24})+…}{x^5}\)
Now, coefficient of x and x³ should be zero and that of x⁵ should be 1, n
? B + c = 0
? ^b⁄₆ + ^c⁄₂ = a
? ^b⁄₁₂₀ + ^c⁄₂₄ = 1
? By solving se 3 equations, a = 1.37, b = 4.13, c = -4.13.

xSin(x) + ayCos(x) + Tan(y) =0
Differentiation of above eqn. is
\(Sin(x) + xCos(x) ? ayCos(x) + aCos(x) \frac{dy}{dx} + Sec^2 (y) \frac{dy}{dx}=0\)
\(\frac{dy}{dx}=\frac{[ayCos(x)-Sin(x)+xCos(x)]}{[aCos(x)+Sec^2 (y)]}\)

^dz⁄_dx = ^d⁄_dx e^x Sin(Cos(x))Cos(Sin(x)) = [(e^x Sin(Cos(x))Cos(Sin(x)) – e^x Sin(x)Cos(Cos(x))Cos(Sin(x)) – e^x Cos(x)Sin(Cos(x))Sin(Sin(x)))].

The problem asks us to find limit of function f(x, y) along two lines y = x and y = 2x
For first line (first person)
x = t : y = 2t
\(=lt_{t\rightarrow 0}\frac{x^{-6}.2^7.t^7}{t+2t}=lt_{t\rightarrow 0}\frac{2^7t}{3t}\)
=\(\frac{2^7}{3}\)
For second line (Second Person)
x = t = y
=\(lt_{t\rightarrow 0}\frac{t^{-6}.t^7}{t+t}=lt_{t\rightarrow 0}\frac{t}{2t}\)
=1/2
The limits are different and y will not meet.

\(\lim_{x\rightarrow 1}[x]cos(\frac{\pi(1-x)}{2})e^{1/(1-x)}\)
We know that
x-1 ≤ [x] ≤ x
Multiplying by Remaining term of question
\((x-1)e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})≤e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})≤[x]≤xe^{1/(1-x)}cos(\frac{\pi(1-x)}{2})\)
\(\lim_{x\rightarrow 1}(x-1)e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})≤\lim_{x\rightarrow 1}e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})[x]\)
\(≤\lim_{x\rightarrow 1}xe^{1/(1-x)}cos(\frac{\pi(1-x)}{2})\)
By rearranging terms of e^1/(1-x) to e^-1/(1-x)
\(\lim_{x\rightarrow 1}e^{-1/(x-1)}cos(\frac{\pi(1-x)}{2})x-1≤\lim_{x\rightarrow 1}e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})[x]\)
\(≤\lim_{x\rightarrow 1}e^{-1/(x-1)}cos(\frac{\pi(1-x)}{2})x\)
\(0≤e^{-1/(x-1)}cos(\frac{\pi(1-x)}{2})[x]≤0\)
Hence by sandwich rule
\(\lim_{x\rightarrow 1}e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})[x]=0\)

Sin(y)=Sin^(-1) (y)
Differentiating both sides
\(Cos(y) \frac{dy}{dx}=\frac{1}{\sqrt{1-y^2}} \frac{dy}{dx}\)
(1-y²)(1-Sin² y)=1

\(\frac{dSin(x)Cos(x)}{dx} = Cos(x) \frac{dSin(x)}{dx} + Sin(x) \frac{dCos(x)}{dx}\) = Cos²(x) ? Sin²(x) = Cos(2x).

wkt \(J(\frac{u,v}{x,y}) = \frac{?(u,v)}{?(x,y)} = \begin{vmatrix}
\frac{?u}{?x} & \frac{?u}{?y}\\
\frac{?v}{?x} & \frac{?v}{?y}\\
\end{vmatrix}\)
But we have to find u and v first using given equation
u+v=e^x cos ?y???(1)
u-v=e^x sin? y??..(2)
solving (1)&(2)
we get u=\(\frac{e^x}{2} \)(cos?y+sin?y)
v=\(\frac{e^x}{2} \)(cos?y-sin?y)
\(\frac{?u}{?x} = \frac{e^x}{2}(cos?y+sin?y) \)
\(\frac{?v}{?x} = \frac{e^x}{2}(cos?y-sin?y) \)
\(\frac{?u}{?y} = \frac{e^x}{2}(cos?y-sin?y) \)
\(\frac{?v}{?y} = \frac{e^x}{2}(-cos?y-sin?y) \)
\(J(\frac{u,v}{x,y}) = \begin{vmatrix}
\frac{e^x}{2}(cos?y+sin?y) & \frac{e^x}{2}(cos?y-sin?y)\\
\frac{e^x}{2}(cos?y-sin?y) & \frac{e^x}{2}(-cos?y-sin?y)\\
\end{vmatrix}\)
\((\frac{e^x}{2})(\frac{e^x}{2})\){-(cos?y+sin?y)² -(cos?y-sin?y)²}??.(expanding & solving by taking cos² y+sin² y=1)
=\(\frac{-e^2x}{2} \).