How do you find the linearization of e^sin(x) esin(x) at x=1? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Konstantinos Michailidis May 26, 2016 The linearization of f(x)=e^sinxf(x)=esinx at x=1x=1 is f(x)~~f(1)+f'(1)*(x-1) Hence f(1)=e^sin1~~2.32 and f'(1)=e^(sin1)*cos1~~1.25 Hence f(x)~~2.32+1.25*(x-1) Answer link Related questions How do you find the linear approximation of (1.999)^4 ? How do you find the linear approximation of a function? How do you find the linear approximation of f(x)=ln(x) at x=1 ? How do you find the tangent line approximation for f(x)=sqrt(1+x) near x=0 ? How do you find the tangent line approximation to f(x)=1/x near x=1 ? How do you find the tangent line approximation to f(x)=cos(x) at x=pi/4 ? How do you find the tangent line approximation to f(x)=e^x near x=0 ? How do you use the tangent line approximation to approximate the value of ln(1003) ? How do you use the tangent line approximation to approximate the value of ln(1.006) ? How do you use the tangent line approximation to approximate the value of ln(1004) ? See all questions in Using the Tangent Line to Approximate Function Values Impact of this question 2769 views around the world You can reuse this answer Creative Commons License