We can express 2^x as e^(ln 2 * x)
Thus,
2^[sin (pi x)]=e^[ln 2 *sin (pi x)]
Now, the chain rule says that,
e^{f(x)} then the derivative is, f'(x)e^{f(x)}
That is take derivate of exponent and bring it down.
In this case the exponent is,
ln 2* sin(pi x) where ln 2 is a constant function,
The derivative is after using chain rule,
pi*ln 2*cos(pi x)
Thus, the answer is,
(pi*ln 2*cos(pi x))e^[ ln 2*sin (pi x)]
But, the 'e' exponent simplifies to (look at beginning of post),
(pi*ln 2*cos(pi x))2^[sin (pi x)]
You use a double chain rule here.
The inner function is,
1+[cos x]^2
The derivative of this function is,
-2 sin x cos x=-sin 2x (double angle)
Thus, derivative of
(1+cos^2 x)^6 is,
Inner function derivative which we found to be,
-sin 2x
Times outer function derivative,
6(1+cos^2 x)^5
Multiply them out,
-6*sin 2x(1+cos^2 x)^5
sorry about using that word, i didnt think it would have been considered inappropriate =/
can you show me how
1+[cos x]^2 = -2 sin x cos x
this whole time i tohught i wasnt getting the chain rule, but this is the part that got me.
the answer given was -12cosxsinx(1+cos^2x)^5
New questions should be posted in new threads!!
Are you asking how to take the derivative of 1 + [cos(x)]^2? Because
1+[cos x]^2 is most certainly NOT -2 sin x cos x. (For starters the LHS is always positive.)
d/dx{1 + [cos(x)]^2} = 0 + d/dx{[cos(x)]^2}
Your "outer function" is y^2 where y = cos(x) and your "inner function" is cos(x), so take the derivative of y^2 (2*y) and multiply that by the derivative of cos(x): ( [f(g(x))]' = f'(g(x))*g'(x) )
d/dx{[cos(x)]^2} = 2*cos(x)* -sin(x) = -2sin(x)cos(x)
-Dan