Determine the derivative of each function
A1) y = e^x(Sin²x)
A2) y = e²Sin²x
B1) y = (x³ + 1)e^-2
B2) y = (x² + 1)e^-2
C) y = x²(theta^Sinx)
I am not going to do all these
if y = u(x) v(x)
then dy/dx = u dv/dx + v du/dx
for A1
u = e^x so du/dx = e^x
v = sin^2 x so dv/dx = 2 sin x cos x
so
dy/dx = 2e^x sin x cos x + e^x sin^2 x
= e^x sin x ( 2 cos x + sin x)
Could you help me for these two:
b) y = (x²+1)e^-4
c) y = x²(e^Sinx)
Please & Thank you.
e^-4 is a constant
so
e^-4(2x)
x^2 d/dx (e^sin x) + 2 x e^sin x
x^2 (e^sin x)cos x + 2 x e^sin x
x e^sin x [ x cos x + 2 ]
The answer for b) is e^-4(2x) ?
And everything after that is for c) right. So the final answer for c) would be xe^sinx(xcosx + 2)
?
Thankyou.
In part c) for xcosx could you simplify that or do you keep it as xcosx? Thanks
To find the derivative of each function, we can use the chain rule and product rule.
A1) y = e^x(Sin²x)
First, let's break down the function into two parts: f(x) = e^x and g(x) = Sin²x.
To find the derivative of f(x), we simply differentiate e^x, which gives us f'(x) = e^x.
To find the derivative of g(x), we use the chain rule. The derivative of Sin²x is 2Sin(x) * Cos(x). So, g'(x) = 2Sin(x) * Cos(x).
Now, we can apply the product rule:
y' = f'(x) * g(x) + f(x) * g'(x)
= e^x * Sin²x + e^x * 2Sin(x) * Cos(x)
= e^x * (Sin²x + 2Sin(x) * Cos(x))
A2) y = e²Sin²x
Here, we have f(x) = e² and g(x) = Sin²x.
The derivative of f(x) is simply 0 because e² is a constant.
To find the derivative of g(x), we use the same result as in A1, which is 2Sin(x) * Cos(x).
Now, we can apply the product rule:
y' = f'(x) * g(x) + f(x) * g'(x)
= 0 * Sin²x + e² * 2Sin(x) * Cos(x)
= 2e² * Sin(x) * Cos(x)
B1) y = (x³ + 1)e^-2
In this case, we have a product of two functions: f(x) = x³ + 1 and g(x) = e^-2.
To find the derivative of f(x), we differentiate x³ + 1, which gives us f'(x) = 3x².
To find the derivative of g(x), we differentiate e^-2, which is simply 0 because e^-2 is a constant.
Now, we can apply the product rule:
y' = f'(x) * g(x) + f(x) * g'(x)
= (3x²) * e^-2 + (x³ + 1) * 0
= 3x² * e^-2
B2) y = (x² + 1)e^-2
Similar to B1, here we have f(x) = x² + 1 and g(x) = e^-2.
The derivative of f(x) is f'(x) = 2x.
Again, the derivative of g(x) is 0 because it's a constant.
Applying the product rule:
y' = f'(x) * g(x) + f(x) * g'(x)
= (2x) * e^-2 + (x² + 1) * 0
= 2xe^-2
C) y = x²(theta^Sinx)
In this case, we have f(x) = x² and g(x) = theta^Sinx.
To find the derivative of f(x), we differentiate x², which gives us f'(x) = 2x.
To find the derivative of g(x), we have to use the chain rule. The derivative of theta^Sinx is theta^Sinx * Cosx * Sinx.
Applying the product rule:
y' = f'(x) * g(x) + f(x) * g'(x)
= (2x) * theta^Sinx + x² * (theta^Sinx * Cosx * Sinx)
= 2x * theta^Sinx + x² * theta^Sinx * Cosx * Sinx
These are the derivatives of the given functions.