Question

How do you find the gradient of the tangent to the curve $y = {x^3}$ at the given value of $x = 4$ ?

Answer 1

To solve this problem we should know about how can we get the gradient of the tangent to the curve: To get tangent of curve we had to calculate the differentiation of that curve which is equal to gradient of the tangent to the curve.
Tangent: The line which touches the curve but does not cross it.
The slope of the tangent line at a point on the function is equal to the derivative of the function at the same point.
Differentiation: It is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables.

Formula Used: $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$

Complete step by step solution:
As given in question $2\sin x\cos x + \cos x = 0$ .
In $2\sin x\cos x + \cos x = 0$ , we take $\cos x$ as common. We get,
$\Rightarrow \cos x(2\sin x + 1) = 0$
$\Rightarrow \cos x = 0$
$\Rightarrow (2\sin x + 1) = 0$
For,
$\Rightarrow \cos x = 0$
Whose solution in domain $[0,2\pi ]$ is \left\\{ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right\\}
$\Rightarrow 2\sin x + 1 = 0$
$\Rightarrow \sin x = - \dfrac{1}{2}$
Whose solution in domain $[0,2\pi ]$ is \left\\{ {\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}} \right\\}
Adding a different solution set for both of them. We get,

Hence, solution set is \left\\{ {\dfrac{\pi }{2},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}} \right\\}

Note: As tangent passes through the point where the tangent line and the curve meet, called the point of tangency, the tangent line is “going in the same direction” as the curve, and is thus the best straight-line approximation to the curve at that point.