Y is equal to the second power of x plus the 3x power of e minus Ln3

Y is equal to the second power of x plus the 3x power of e minus Ln3

The result is: 2x + 3E 3x power, input method is not easy to adjust, good luck, hope to adopt

Let the derivative of function f be f ', and F = f'sinx + cosx Then f '=

f=f'sinx+cosx
f'(x)=f'(π/2)cosx-sinx
When x = π / 2,
f'(π/2)=f'(π/2)cosπ/2-sinπ/2=-1
f'(x)=-cosx-sinx
∴f'(π/4)=-cosπ/4-sinπ/4
=-√2/2-√2/2=-√2

What is the first derivative of SiNx to the third power

[(sinx)^3]'=3[sinx^2]cosx

What is the n-th derivative of SiNx to the third power?

(sinx)^3＝3/4×sinx－1/4×sin3x
The n-th derivative of (SiNx) ^ 3 = 3 / 4 × sin (x + n π / 2) - 1 / 4 × 3 ^ n × sin (3x + n π / 2)

What's the derivative of (SiNx) to the third power

∫（sinx)^3dx=-∫（sinx)^2dcosx
=-∫（1-cosx^2)dcosx
=1 / 3 (cosx ^ 3) - cosx + C C C is an arbitrary constant
In other words, the derivative of 1 / 3 (cosx ^ 3) - cosx + C is the third power of (SiNx)

The derivative of SiNx to the fourth power

4 * (the third power of SiNx) * cosx

Find the derivative of 2 to the SiNx power!

=SiNx power of 2 × in2 × (SiNx) '
=SiNx power of LNX cosx 2

Let's find the derivative of y = SiNx to the third power, the derivative of sin ^ 3x, where 3 is in the upper right corner of sin,

y=sin³x
y'=(3sin²x)×(sinx)'
y'=3cosxsin²x

Find the derivative of function, y = e to the power of X SiNx / X

y=e^x(sinx/x)
be
y'=(e^x)'(sinx/x)+e^x(sinx/x)'
=e^x(sinx/x)+e^x((xcosx-sinx)/x^2)

The derivative of the power X of y = [x / (1 + x)] by logarithmic derivation

y=[x/(1+x)]^x
lny=x*ln(x/(1+x))
y'/y=[x*ln(x/(1+x))]'
y'/y=ln(x/(1+x))+x*[ln(x/(1+x))]'
y'/y=ln(x/(1+x))+x*[(1/x)+1/(1+x)]
y'=y*{ln(x/(1+x))+x*[(1/x)+1/(1+x)]}
=[x/(1+x)]^x*(lnx-ln(1+x)+x/(1+x)+1)