How do I find dx/dt implicitly with the function $\tan(x)=4t^{-1}$? This question is a trigonometric one from what I can tell. I understand that you need to differentiate the given function and the unknown variable on top ($dx$) will correspond to the $x$ unknowns, for example $5x^2$ will give us $10x\frac{dx}{dt}$.
With this specific question I can only get to $-\frac{\sec^2x}{4t^2}$ but the answer is $-\frac{\cos^2x}{4t^2}$. I also do not understand how $\sec^2x$ becomes $\cos^2x$.
Thanks.
 A: $$\frac{d(\tan(x))}{dt}=\sec^2x \frac{dx}{dt}$$
$$\frac{d(4t^{-1})}{dt}=-4t^{-2}$$
$$\frac{dx}{dt}=\frac{-4t^{-2}}{\sec^2x}=-\frac{4\cos^2x}{t^2}\because \cos^2x=\frac{1}{\sec^2x}$$
A: I guess x is a function of t
$$\tan(x)=4t^{-1}$$
$$\frac {d\tan(x)}{dt}=-4t^{-2}$$
$$\frac {x'}{\cos^2(x)}=-4t^{-2}$$
$$ x'= \frac {dx}{dt}=-4{\cos^2(x)}t^{-2}$$
Note that $\sec(x) =\frac 1 {\cos(x)}$
A: If you find implicit differentiation a bit confusing, try considering your variables to be functions of some third variable, such as $s$ and take the derivative with respect to that variable.
\begin{eqnarray}
\tan x&=&4t^{-1}\\
\frac{d}{ds}\tan x&=&\frac{d}{ds}4t^{-1}\\
\sec^2x\frac{dx}{ds}&=&-4t^{-2}\frac{dt}{ds}\\
\sec^2x\,dx&=&-4t^{-2}\,dt\\
\frac{dx}{dt}&=&-\frac{4t^{-2}}{\sec^2x}\\
&=&-\frac{4\cos^2x}{t^2}
\end{eqnarray}
A: Rewrite the relation as
$$t\,\tan x=4,\quad\text{so }\tan x=\frac 4t,$$
and differentiate this relation:
\begin{align}
&&&\mathrm d t\,\tan x+t(1+\tan^2x)\,\mathrm d x=0 \\[1ex]
&\text{whence}\qquad&\frac{\mathrm d x}{\mathrm d t}&=-\frac{\tan x}{t(1+\tan^2x)}=-\frac{t\tan x}{t^2+(t\tan x)^2}\\[1ex]
&&&=\color{red}{\frac 4{t^2+16}}.
\end{align}
A: The easiest way to do implicit derivatives is to separate differentiation from the derivative.  First, differentiate, then solve for your derivative.
Example: $y = x^2$.  Differential of $y$ is $dy$.  Differential of $x^2$ is $2x\,dx$.  Therefore $dy = 2x\,dx$.  Solving for the derivative gives $\frac{dy}{dx} = 2x$.
So, to your problem:
$$ tan(x) = 4t^{-1} \\
d(tan(x)) = d(4t^{-1}) \\
sec^2(x)\,dx = -4t^{-2}\,dt \\
$$
Now just solve for $\frac{dx}{dt}$:
$$\frac{dx}{dt} = \frac{-4t^{-2}}{sec^2(x)}$$
Or, more simply:
$$\frac{dx}{dt} = -4\frac{cos^2(x)}{t^2} $$
