How do you find the equation of a line tangent to the function #y=4/sqrtx# at (1,4)?

Question

Christopher Allers · Accepted Answer

#y=-2x+6

The equation of the tangent in #color(blue)"point-slope form"# is. #color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))# where m represents the slope and #(x_1,y_1)" a point on the line"# #color(orange)"Reminder"color(white)(xxx)m=dy/dx# Express #y=4/sqrtx=4/x^(1/2)=4x^(-1/2)# #rArrdy/dx=-2x^(-3/2)=-2/x^(3/2)# #x=1:dy/dx=-2/1^(3/2)=-2/1=-2=m# substitute m = - 2 and #(x_1,y_1)=(1,4)# into the equation. #y-4=-2(x-1)rArry-4=-2x+2# #rArry=-2x+6" is the equation of the tangent"#

Isabella Ackerman · Answer

undefined To find the equation of a line tangent to the function y = 4/√x at the point (1,4), we can use the concept of differentiation.

First, we need to find the derivative of the function y = 4/√x.

Using the power rule for differentiation, the derivative of y = 4/√x can be found as follows:

dy/dx = d/dx (4/√x) = -4/(2√x) = -2/√x

Next, we substitute the x-coordinate of the given point (1,4) into the derivative to find the slope of the tangent line at that point.

Substituting x = 1 into dy/dx, we get:

dy/dx = -2/√1 = -2

Therefore, the slope of the tangent line at the point (1,4) is -2.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as:

y - y1 = m(x - x1)

Substituting the values of (x1, y1) = (1,4) and m = -2, we have:

y - 4 = -2(x - 1)

Simplifying the equation, we get:

y - 4 = -2x + 2

Finally, rearranging the equation to the standard form, we have:

2x + y = 6

Therefore, the equation of the line tangent to the function y = 4/√x at the point (1,4) is 2x + y = 6.