How do you differentiate #f(x)=(sinx+1)(x^2-3e^x)# using the product rule?

Question

Emma Aldridge · Accepted Answer

#\cos (x)(x^2-3e^x)+(2x-3e^x)(\sin (x)+1)# #\frac{d}{dx}((\sin (x)+1)(x^2-3e^x))# Applying product rule, #(f\cdot g)^'=f^'\cdot g+f\cdot g^'# #f=sinx +1 ,g=x^2 - 3e^x# #=frac{d}{dx}(sin (x)+1)(x^2-3e^x)+frac{d}{dx}(x^2-3e^x)(sin (x)+1)# we know, #\frac{d}{dx}(sin (x)+1)=cos (x)#; #\frac{d}{dx}(\sin (x))=\cos(x)#; #\frac{d}{dx}(1)=0#; #\frac{d}{dx}(x^2-3e^x)=2x-3e^x# Finally, #=\cos(x)(x^2-3e^x)+(2x-3e^x)(\sin (x)+1)#

Aurora Alvarado · Answer

undefined To differentiate $ f(x) = (\sin x + 1)(x^2 - 3e^x) $ using the product rule, follow these steps:

1. Identify the functions $ u(x) $ and $ v(x) $.
2. Apply the product rule formula: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
3. Differentiate each function $ u(x) $ and $ v(x) $ separately.
4. Substitute the derivatives and original functions into the product rule formula.
5. Simplify the expression.

Here's the step-by-step process:

1. Let $ u(x) = \sin x + 1 $ and $ v(x) = x^2 - 3e^x $.
2. Differentiate $ u(x) $ and $ v(x) $ separately:

$ u'(x) = \cos x $ (derivative of $ \sin x $)  
   $ v'(x) = 2x - 3e^x $ (derivative of $ x^2 - 3e^x $)

3. Apply the product rule formula:

$ f'(x) = (\sin x + 1)(2x - 3e^x) + (\cos x)(x^2 - 3e^x) $

4. Simplify the expression:

$ f'(x) = 2x\sin x - 3e^x\sin x + 2x - 3e^x + x^2\cos x - 3e^x\cos x $

So, the derivative of $ f(x) $ is $ f'(x) = 2x\sin x - 3e^x\sin x + 2x - 3e^x + x^2\cos x - 3e^x\cos x $.