How do I find dx/dt implicitly with the function $\tan(x)=4t^{-1}$?

Question

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.

Answer 1

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}

Answer 2

$$\frac{d(\tan(x))}{dt}=\sec^2x \frac{dx}{dt}$$ $$\frac{d(4t^{-1})}{dt}=-4t^{-2}$$ $$\frac{dx}{dt}=\frac{-4t^{-2}}{\sec^2x}=-\frac{4\cos^2x}{t^2}\because \cos^2x=\frac{1}{\sec^2x}$$

Answer 3

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 $\sec(x) =\frac 1 {\cos(x)}$

Answer 4

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}

Answer 5

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: $$\frac{dx}{dt} = -4\frac{cos^2(x)}{t^2} $$