#### To determine

**a)**

**To State:** The derivative of y=ex.

#### Answer

ex.

#### Explanation

**Calculation:**

**Given:**y=ex.

ddxex=limh→0ex+h−exh

By laws of exponents, one can split the addition of exponents into multiplication of the same base.

ddxex=limh→0exeh−exh=limh→0ex(eh−1)h=exlimh→0eh−1h

Since, limx→0ex−1x=f′(0)=1

So, ddxex=ex

**Final statement:**

Hence, derivative of y=ex is ex.

#### To determine

**b)**

**To State:** The derivative of y=bx.

#### Answer

bx×lnb.

#### Explanation

**Calculation:**

**Given:** y=bx.

This allow to differentiate the function using the chain rule:

ddx(exlnb)=exlnb×ddx[xlnb]

Just like ddx[5x]=5, ddx(xlnb)=lnb, since lnb will always be constant.

So, ddx(bx)=bxlnb

**Final statement:**

Hence, the derivative of y=bx is equal to bx×lnb.

#### To determine

**c)**

**To State:** The derivative of y=lnx.

#### Answer

1x.

#### Explanation

**Calculation:**

**Given:**y=lnx.

Derivative of lnx is 1x, so the constant multiple is -1 has no effect on derivative and comes with the answer i.e. =1x.

**Final statement:** Hence, the derivative of y=lnx is equal to 1x.

#### To determine

**d)**

**To State:** The derivative of y=logbx.

#### Answer

1x1lnx.

#### Explanation

**Calculation:**

**Given:** y=logbx.

That depends on what base one intend.logx is sometimes used for log10x, logex and log2x.

So, ddx(logbx)=1x×1logex

Since lnx=logex

Therefore

ddx(logbx)=1x1lnx

**Final statement:** Hence, the derivative of y=logbx is equal to 1x1lnx.

#### To determine

**e)**

**To State:** The derivative of y=sin−1x.

#### Answer

11−x2.

#### Explanation

**Calculation:**

**Given:**y=sin−1x.

Let us derive the derivative:

Let y=sin−1x

By rewritting in terms of sine

** **siny=x

by implicitly differentiating with respect to x:

cosydydx=1

By dividing by cosy:

dydx=1cosy

by cosy=1−sin2y

dydx=11−sin2y

by siny=x

So,dydx=11−x2

**Final statement:** Hence, the derivative of y=sin−1x is 11−x2.

#### To determine

**f)**

**To State:** The derivative of y=cos−1x.

#### Answer

−11−x2.

#### Explanation

**Calculation:**

**Given:** y=cos−1x.

Let y=cos−1x

cosy=xddxcosy=ddx(x)−sinydydx=1dydx=−1siny

Now apply the identity from trigonometry:

siny=±1−cos2y

So,

dydx=−11−cos2yx=cosydydx=−11−x2

**Final statement:**

Hence, the derivative of y=cos−1x is −11−x2.

#### To determine

**g)**

**To State:** The derivative of y=tan−1x.

#### Answer

**Solution**: 11+x2.

#### Explanation

**Calculation:**

**Given:**y=tan−1x.

Let y=arc tanx

tany=xsec2ydydx=1dydx=1sec2y

Since tany=x1 and 12+x2=1+x2

sec2y=(1+x21)2=1+x2

dydx=11+x2

**Final statement:**

Hence, the derivative of y=tan−1x is 11+x2.

#### To determine

**h)**

**To State:** The derivative of y=sinhx.

#### Answer

coshx.

#### Explanation

**Calculation:**

**Given:** y=sinhx.

sinhx=ex−e−x2coshx=ex+e−x2

Differentiate from here using either the quotient rule or the sum rule.

By sum rule:

sinhx=ex−e−x2=ex2−e−x2

ddxsinhx=ddx(ex2−e−x2)=ex2−(−e−x2)=ex+e−x2=coshxddxsinhx=coshx

**Final statement:**

Hence, the derivative of y=sinhx is coshx.

#### To determine

**i)**

**To State:** The derivative of y=coshx.

#### Answer

sinhx.

#### Explanation

**Calculation:**

**Given:** y=coshx.

A given value but the proof is that:

sinhx=ex−e−x2coshx=ex+e−x2

And derivative is:

12(ddx)ex+12(ddx)e−x

The derivative of any ex is ex also the −x brings down the negative because of the chain rule.

get 12ex+12e−x(−1)

=12(ex−e−x)ddxcoshx=sinhx

**Final statement:**

Hence, the derivative of y=coshx is sinhx.

#### To determine

**j)**

**To State:** The derivative of y=tanhx.

#### Answer

1−tanh2x.

#### Explanation

**Calculation:**

**Given:**y=tanhx.

Since,

tanhx=ex−e−xex+e−xddxtanhx=(ex+e−x)(ex+e−x)−(ex−e−x)(ex−e−x)(ex+e−x)2=1−(ex−e−xex+e−x)2ddxtanhx=1−tanh2x

**Final statement:**

Hence, the derivative of y=tanhx is 1−tanh2x.

#### To determine

**k)**

**To State:** The derivative of y=sinh−1x.

#### Answer

**Solution**: 1x2+1.

#### Explanation

**Calculation:**

**Given:**y=sinh−1x.

Since,

sinh−1x=ln(x+x2+1)

So, ddxsinh−1x=ddx(ln(x+x2+1))

=1x+x2+1×(1+12x2+1⋅2x)=1x+x2+1×(x2+1x2+1+xx2+1)=1x+x2+1×((x2+1+x)x2+1)=1x2+1

**Final statement:**

Hence, the derivative of y=sinh−1x is 1x2+1.

#### To determine

**l)**

**To State:** The derivative of y=cosh−1x.

#### Answer

1x2−1.

#### Explanation

**Calculation:**

**Given:**y=cosh−1x.

y=cosh−1x

By defination of an inverse function,

x=coshy=ey+e−y2(eyey)x=e2y+12ey

2eyx=e2y+1e2y−2xey+1=0(ey)2−2x(ey)+1=0

ey=2x+4x2−42=x+x2−1ln(ey)=ln(x+x2−1)y=ln(x+x2−1)

cosh−1x=ln(x+x2−1)

So, f′(x)=1x2−1

**Final statement:**

Hence, the derivative of y=cosh−1x is 1x2−1

#### To determine

**m)**

**To State:** The derivative of tanh^{-1} x

#### Explanation

From earlier derivations we will get 11- x2