What is the derivative of #f(x) = xcos^3(x^2)+sin(x)#?

Question

Emma Aldridge · Accepted Answer

#cos^3(x^2) - 6x^2cos^2(x^2)sin(x^2) + cos(x)# Firstly we use the product rule, #(xcos^3(x^2))' = x'cos^3(x^2)+x(cos^3(x^2))'# Now we must use the chain rule on the second term. #(cos^3(x^2))' = 3(cos^2(x^2))(-sin(x^2))(2x))# #:. = (1)cos^3(x^2) + x*3(cos^2(x^2))(-sin(x^2))(2x))# #:. = cos^3(x^2) - 6x^2cos^2(x^2)sin(x^2)# lastly the derivative of #sin(x) = cos(x)# Thus, the complete derivative is #cos^3(x^2) - 6x^2cos^2(x^2)sin(x^2) + cos(x)#

Isabella Ackerman · Answer

undefined To find the derivative of $ f(x) = x\cos^3(x^2) + \sin(x) $, we'll use the sum rule and the chain rule.

The derivative of $ x\cos^3(x^2) $ with respect to $ x $ involves two parts:

1. Differentiating $ x $ with respect to $ x $, which gives $ 1 $.
2. Applying the chain rule to differentiate $ \cos^3(x^2) $. This involves differentiating $ \cos^3(u) $ with respect to $ u $, where $ u = x^2 $, and then multiplying by the derivative of $ x^2 $ with respect to $ x $.

So, the derivative of $ x\cos^3(x^2) $ with respect to $ x $ is $ 1 \cdot \cos^3(x^2) - 3x^2\sin(x^2)\cos^2(x^2) $.

The derivative of $ \sin(x) $ with respect to $ x $ is $ \cos(x) $.

Putting it all together, the derivative of $ f(x) $ is:

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