# `y = root(4)(x), (1, 1)` Find an equation of the tangent line to the curve at the given point.

### Textbook Question

Chapter 3, 3.1 - Problem 33 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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Note:- If f(x) = x^n ; where n = real number, then

f'(x) = n*x^(n-1)

Given equation of curve :-

y = x^(1/4)

Now, the slope of the tangent to the curve at a point (a,b) is given by the value of y'(x) at that point.

Hence, differentiating the function y(x) w.r.t 'x' we get

y'(x) = (1/4)*x^(-3/4)

or, slope of the tangent at the point (1,1) is

y(1) = 1/4

Now, equation of line passing through a point (a,b) and having slope 'm' is :-

y - b = m*(x-a)

Thus, the equation of the tangent at the point (1,1) is :-

y - 1 = (1/4)*(x-1)

or, 4y - 4 = x - 1

or, x - 4y = - 3 = equation of tangent to the given curve at (1,1)

kspcr111 | In Training Educator

Posted on

Given curve is `y=x^(1/4)`

` (dy)/(dx)=(1/4)*x^(-3/4) `

we know ,

slope of the tangent = derivative of the equation of the curve

So the slope of the tangent at (1,1) is given as follows

`((dy)/(dx))_((1,1))=(1/4)*(1)^(-3/4) = (1/4)`

So,the equation of tangent is given as follows

y-1=(1/4)(x-1)

=> y = (1/4)x - (1/4) + 1

=> y = (1/4)x +(3/4)

The graph is as follows