Find dy/dx where y= x^3(sin(x))
To differentiate y, we must used the product rule.The product rule is d/dx [f(x)g(x)] = f'(x)g(x) + g'(x)f(x)So here, we let f(x)= x^3 and g(x)= sin(x)Then, f'(x)= 3x^2 and g'(x) = cos(x)Then substituting these into the product rule formula, we get dy/dx = (3x^2)sin(x) + cos(x)x^3We can simplify the answer by factorising out x^2 :dy/dx= x^2[3sin(x) + xcos(x)]
KC
Related Maths A Level answers
All answers ▸Prove that the equation y = 3x^4 - 8x^3 - 3 has a turning point at x=2
The quadratic equation x^2 + 4kx+2(k+1) = 0 has equal roots, find the possible values of k.
Using logarithms solve 8^(2x+1) = 24 (to 3dp)
Find the stationary point(s) of the curve: y = 3x^4 - 8x^3 - 3.