(x)=√(5x+10), find f′(x) using the definition f′(x)= lim, h→0, {f(x+h)−f(x)} / h . Then, using this, find the tangent line to the graph of y= √(5x+10) at x=3. Enter

Question

For f(x)=√(5x+10), find f′(x) using the definition f′(x)= lim, h→0, {f(x+h)−f(x)} / h . Then, using this, find the tangent line to the graph of y= √(5x+10) at x=3. Enter all values as integers, or fractions in lowest terms.

Answer 1

f(x) = √(5x + 10)

f(x+h) = √(5(x+h) + 10)

From the limit definition

f '(x) = lim _h→0 [f(x + h) - f(x)] / h

= lim _h→0 [√(5(x+h) + 10) - √(5x + 10)] / h

We can simplify this by multiplying the top and bottom by the conjugate √(5(x+h) + 10) + √(5x + 10)

We get:

f ' (x) = lim _h→0 [(5(x+h) + 10) - (5x + 10)] / (h * [√(5(x+h) + 10) + √(5x + 10)] )

= lim _h→0 (5h) / (h * [√(5(x+h) + 10) + √(5x + 10)] )

= lim _h→0 5 / [√(5(x+h) + 10) + √(5x + 10)] )

Now we can plug 0 in for h

f '(x) = 5 / [√(5x + 10) + √(5x + 10)]

= 5 / (2√(5x + 10))

Now to find the tangent line at x = 3, get the slope of the tangent line which is just f '(3). Derivatives tell you the slopes of tangent lines at different values of x.

Then get a point that the tangent line goes through. The tangent line intercepts the function at x = 3.

Therefore, a point that it goes through is (3, f(3))

Lastly, use the point-slope form to get the equation of the line.