f'(x) = e^x

f'(x) = xe^x + xe^x

f'(x) = e^x - xe^x

f'(x) = e^x + xe^x

a

b

c

d

 $\frac{1}{\cos\left(5x^2\right)}$  

 $-\frac{\left(10x\ \sin\left(5x^2\right)\right)}{\ \cos\left(5x^2\right)}$  

 $-\frac{1}{10x\sin\left(5x^2\right)}$  

 $\frac{-\sin\left(5x^2\right)}{\cos\left(5x_{ }^2\right)}$  

y′=1 /((log9)x)

y′=(ln9)/x

y′=1/(x(lnx))

y′=1/((ln9)x)

A

B

C

D

h^'(x)=((g(x)f^'(x) + f(x)g^'(x)) / (g(x))²

h^'(x)=((g^'(x)f^'(x) - f(x)g(x)) / (g(x))

h^'(x)=((g(x)f^'(x) - f(x)g^'(x)) / (g(x))²

h^'(x)=((g^'(x)f^'(x) + f(x)g(x)) / (g(x))

h^'(x)=(f^'(g(x))(f^'(x))

h^'(x)=(f^'(g(x))(g(x))

h^'(x)=(f^'(g(x))(f(x))

h^'(x)=(f^'(g(x))(g^'(x))

position

velocity

calculus

particle motion