Question

How to solve to find the equation of the tangent line to the curve $f=y=(x+1)\cos x$ at the given point (0,1)?

Answer 1

To solve the given question we need to know the concept of equation of straight line. The tangent of a curve at any point is always a straight line which is $y=mx+c$ where $x,y$ are the coordinates,$m$ is the slope of the tangent at the given point and $c$ is the constant. We also need to know the concept of differentiation, as $m=\dfrac{dy}{dx}$.

Complete step-by-step answer:
The function in the the given problem is $y=(x+1)\cos x$ . The equation of the tangent is$y=mx+c$. Firstly we need to find the value of “m” which is the slope of the tangent at (0,1).
To know the value of slope at (0,1) we are suppose to differentiate the given curve, first .
$\dfrac{dy}{dx}=\dfrac{d((x+1)\cos x)}{dx}$
On applying the distributive property in the above equation :
$\Rightarrow \dfrac{d(x\cos x+\cos x)}{dx}$
$\Rightarrow \dfrac{d(x\cos x)}{dx}+\dfrac{d(\cos x)}{d}$
For differentiating we will use the formula of differentiation of $u\times v$which will result in $u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ ,here $u$ and $v$ are the function in terms of $x$
$\Rightarrow \dfrac{d(x\cos x)}{dx}=x\dfrac{d\cos x}{dx}+\cos x\dfrac{dx}{dx}$
On differentiating the above equation
$\Rightarrow x(-\sin x)+\cos x(1)$
Therefore on differentiation of the curve
$\Rightarrow x(-\sin x)+\cos x(1)-\sin x$
On putting the value of $x=0$ in the above equation we get
$\Rightarrow 0(-\sin 0)+\cos 0(1)-\sin 0$
$\because \sin 0=0$ and $\cos 0=1$ the slope turns to be
$\begin{aligned} & \Rightarrow 0+1-0 \\\ & \Rightarrow 1 \\\ \end{aligned}$
The slope of the curve at $x=0$ is 1.
Now the equation of the tangent $y=mx+c$, where m=1 . Lets put the value of each term in the given equation. So as per the question $y=1$ and $x=0$ is given and we have found the value of$m=1$ .
$y=mx+c$
On putting the values
$\Rightarrow 1=1\times 0+c$
$\Rightarrow 1=c$
Thus the value of $c$ becomes $1$.
$\therefore$ The equation of the tangent is $y=x+1$

Note: Always take care of the calculation. The slope on the curve can be positive or negative , so you need to put the value of the $m$ with the sign. Do remember the formulas of differentiation. Remember the slope of the tangents at different points on the same curve is different. So slope of the tangent at point (2,3) will be different to (1,2) for the same curve .