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How do you differentiate f(x)=e^x(x+2)^2 using the product rule? 12 Dec 2015 · How do you differentiate #f(x)=e^x(x+2)^2# using the product rule? Calculus Basic Differentiation Rules Product Rule. 1 Answer
How do you differentiate f (x)=e^ (3x)cos2x using the ... - Socratic 10 May 2016 · How do you differentiate #f(x)=e^(3x)cos2x# using the product rule? Calculus Basic Differentiation Rules Product Rule. 1 Answer
What is the derivative of #y=(sinx)^(2x)#? - Socratic 2 Mar 2017 · What is the derivative of #f(x)=(log_6(x))^2# ? See all questions in Differentiating Logarithmic Functions without Base e Impact of this question
How do you differentiate f (x)=e^ (x^2)/ (e^ (2x)-2x) using the ... 30 Nov 2015 · How do you differentiate #f(x)=e^(x^2)/(e^(2x)-2x)# using the quotient rule? Calculus Basic Differentiation Rules Quotient Rule. 1 Answer
How do you implicitly differentiate #2= e^(xy^2-xy)-y^2x^3+y 3 Apr 2017 · How do you find the second derivative by implicit differentiation on #x^3y^3=8# ? What is the derivative of #x=y^2#? See all questions in Implicit Differentiation
How do you find the derivative of #(x-3) /( 2x+1)#? - Socratic 16 Jun 2018 · How do I find the derivative of #y= x arctan (2x) - (ln (1+4x^2))/4#? How do you find the derivative of #y = s/3 + 5s#? What is the second derivative of #(f * g)(x)# if f and g are …
How do you implicitly differentiate #2= e^(xy^2-x^3y)-y^2x^3+y 22 Apr 2018 · How do you Use implicit differentiation to find the equation of the tangent line to the curve
What is the derivative of #y=e^(3-2x)# - Socratic 7 Sep 2014 · By Chain Rule, y'=-2e^{3-2x}. Recall: Exponential Rule: (e^x)'=e^x Chain Rule: [f(g(x))]'=f'(g(x))cdot g'(x) Since f(x)=e^x and g(x)=3-2x, we have y'=e^{3-2x}cdot(-2 ...
How do you differentiate g(x) = 2xe^(2x) using the product rule ... 3 Mar 2016 · f=2x,g=e^(2x) ->f'=2,g'=2e^(2x) g'(x)=fg'+gf'=4xe^(2x)+2e^(2x) Separate the products into f and g then find each of their derivative before putting it into the product rule
How do you find the first derivative of #e^(x^2)#? - Socratic (dy)/(dx)=2xe^(x^2) Right now, you have y=e^(x^2) The derivative of y=e^(f(x)) is (dy)/(dx)=f'(x)e^(f(x)) In this case, f(x)=x^2, and the derivative of x^2=2x ...