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EXAMPLE 6Finding an Equation of a Tangent Line

Find an equation of the tangent line to the graph of f(x)=xx2+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=lnxx2+31+x=lnx+12ln(x2+3)ln(1+x)

Step 2 Differentiate implicitly. yy=1x+122xx2+311+x=1x+xx2+311+x

Step 3 Solve for y and simplify: y=f(x)=y(1x+xx2+311+x)=xx2+31+x(1x+xx2+311+x)

Now we find the slope of the tangent line by evaluating f(1). f(1)=42(1+1412)=34

Then an equation of the tangent line to the graph of f at the point (1,1) is y1=34(x1)y=34x+14