Question

Find the equation of the tangent through a line y = 5x⁴ + 9x⁶ =0 at ( 1,2).
Hint : Use derivatives ​

Answer 1

EXPLANATION.

Equation of tangent through a line.

⇒ y = 5x⁴ + 9x⁶ at (1,2).

As we know that,

Differentiate the equation w.r.t. x, we get.

⇒ dy/dx = 20x³ + 54x⁵.

Put the value of (x, y) = (1, 2) in the equation, we get.

⇒ dy/dx = 20(1)³ + 54(1)⁵.

⇒ dy/dx = 20 + 54.

⇒ dy/dx = 74.

As we know that,

dy/dx is the slope of the equation = m.

⇒ m = 74.

As we know that,

Formula of equation of tangent.

⇒ (y - y₁) = m(x - x₁).

Put the values in the equation, we get.

⇒ (y - 2) = 74(x - 1).

⇒ y - 2 = 74x - 74.

⇒ 74x - 74 - y + 2 = 0.

⇒ 74x - y - 72 = 0.

                                                                                                                         

MORE INFORMATION.

Equation of tangent.

(1) = Equation of tangent to the curve y = f(x) at p(x₁, y₁) is (y - y₁) = m(x - x₁).

(2) = The tangent at (x₁, y₁) is parallel to x-axis : dy/dx = 0.

(3) = The tangent at (x₁, y₁) is parallel to y-axis : dy/dx = ∞.

(4) = The tangent line makes equal angles with the axis : dy/dx = ± 1.

Answer 2

Given,

$\rm \: y = 5 {x}^{4} + 9 {x}^{6}$

we have to find,

The equation of the tangent through a line y = 5x⁴ + 9x⁶ at ( 1,2).

So at first we have to find the slope of the tangent.

We will get the slope by doing 1st derivatives and putting the point on 1st derivatives.

Now,

$\rm \frac{dy}{dx} = \frac{d}{dx} (5 {x}^{4} + 9 {x}^{6} ) \\ \\ = \rm20 {x}^{3 } + 54 {x}^{5} \\ \\ \rm \: slope \: \: m = \frac{dy}{dx} { \huge{ | }} _{(1,2)} = 20 \times {1}^{3} + 54 \times {1}^{5} \\ \\ = > \rm \: m = 74$

So the equation of tangent which slope m =74 and passes through the point (1,2) is :

${ \boxed{\boxed{\begin{array}{cc} \rm \: y - 2 = 74(x - 1) \\ \\ \rm = > y - 2 = 74x - 74 \\ \\ \rm \: = > \boxed{ \rm 74x - y - 72 = 0}\end{array}}}}$