Question

calc help, Derivative of G(x) = sec^2 (xe^(x^(2)))

Answer 1

Put
\[
u = x e^{x^2}
\]

Answer 2

yes d/dx(\sec^2(e^(x^2) x))
Use the chain rule, d/dx(\sec^2(e^(x^2) x)) = ( du^2)/( du) ( du)/( dx), where u = \sec(e^(x^2) x) and ( du^2)/( du) = 2 u: = | 2 \sec(e^(x^2) x) (d/dx(\sec(e^(x^2) x))) | Use the chain rule, d/dx(\sec(e^(x^2) x)) = ( dsec(u))/( du) ( du)/( dx), where u = e^(x^2) x and ( dsec(u))/( du) = \sec(u) \tan(u): = | 2 \sec(e^(x^2) x) (\tan(e^(x^2) x) \sec(e^(x^2) x) (d/dx(e^(x^2) x))) | Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = e^(x^2) and v = x: = | 2 \tan(e^(x^2) x) \sec^2(e^(x^2) x) (e^(x^2) (d/dx(x))+x (d/dx(e^(x^2)))) | Use the chain rule, d/dx(e^(x^2)) = ( de^u)/( du) ( du)/( dx), where u = x^2 and ( de^u)/( du) = e^u:= | 2 \tan(e^(x^2) x) \sec^2(e^(x^2) x) (x (e^(x^2) (d/dx(x^2)))+e^(x^2) (d/dx(x)))
The derivative of x is 1 = 2 \tan(e^(x^2) x) \sec^2(e^(x^2) x) (e^(x^2) x (d/dx(x^2))+e^(x^2)) | The derivative of x^2 is 2 x: = | 2 (e^(x^2) x (2 x)+e^(x^2)) \tan(e^(x^2) x) \sec^2(e^(x^2) x)\]

FINALLY
d/dx(sec^2(x e^(x^2))) = 2 e^(x^2) (2 x^2+1) tan(e^(x^2) x) sec^2(e^(x^2) x)

Answer 3

Power rule for Differentiation:
for:
\[f(x) = x^n\]
then:
\[f'(x)=\frac{d}{dx} n * x^{n-1)} \]
Basically bring the exponent down as a coefficient, then subtract the exponent by one.

Chain Rule:
for:
\[u^n \]
then:
\[(n*u^{n-1})*\frac{d}{du}u\]
Do the power rule, then take the derivative of what's inside the "shell"/"layer"/"whatever"

Answer 4

sec^2 (xe^(x^(2)))

\[
G'[x] = 2 \sec (u) (\sec(x) \tan(x) )u'(x)
\]

Answer 5

writen so far 2tan(xe^(x^(2)))sec^2(xe^(x^(2))) * 1e^(x^(2))+x * ?

Answer 6

\[
G'[x] = 2 \sec (u) (\sec(x) \tan(x) )u'(x)=\\
2 \sec (x e^{x^2}) (\sec(x) \tan(x) )( e^{x^2} + 2 x^2 e^{x^2})
\]