could anyone please tell me how to differentiate those equation;
r(t) = ln (t^2 + 1)i + (t - 2 tan^-1 t)j + 2 (surd2) k
i note that to differentiate ln (t^2 + 1)i = 1/t^2 +1 . (2t) how bout j?
Hello, noob!
Have your forgotten your differentiation formulas?
Could anyone please tell me how to differentiate this equation?
. . $\displaystyle r(t) \:= \:\ln(t^2 + 1)i + (t - 2\tan^{-1}\!t)j + 2\sqrt{2}k$
$\displaystyle f(t)\:=\:\ln(t^2+1) \quad\Rightarrow\quad f'(x) \:=\:\frac{2t}{t^2+1}$
$\displaystyle g(t) \:=\:t - 2\tan^{-1}\!t \quad\Rightarrow\quad g'(t) \:=\:1 - \frac{2}{1 + t^2}$
$\displaystyle h(x) \:=\:2\sqrt{2} \quad\Rightarrow\quad h'(x) \:=\:0$
$\displaystyle y=tan^{-1}(x) \iff \tan(y)=x$
Now taking the implict derivative with respect to x we get
$\displaystyle sec^2(y)\cdot \frac{dy}{dx}=1 \iff \frac{dy}{dx}=\frac{1}{\sec^2(y)}$
$\displaystyle \frac{dy}{dx}=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^ 2(y)}=\frac{1}{1+x^2}$
The last step comes from the eqaution at the top $\displaystyle \tan(y)=x$