Find an equation of the tangent line to the graph of f(x)=x√x2+31+x at the point (1,1).
Solution The slope of the tangent line to the graph of y=f(x) at the point (1,1) is f′(1). Since the function consists of a product, a quotient, and a power, we follow the steps for logarithmic differentiation:
Step 1 Take the natural logarithm of each side, and simplify: lny=lnx√x2+31+x=lnx+12ln(x2+3)−ln(1+x)
Step 2 Differentiate implicitly. y′y=1x+12⋅2xx2+3−11+x=1x+xx2+3−11+x
Step 3 Solve for y′ and simplify: y′=f′(x)=y(1x+xx2+3−11+x)=x√x2+31+x(1x+xx2+3−11+x)
Now we find the slope of the tangent line by evaluating f′(1). f′(1)=√42(1+14−12)=34
Then an equation of the tangent line to the graph of f at the point (1,1) is y−1=34(x−1)y=34x+14